五味子

小舟从此逝,江海寄余生

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简单搜索练习

kuangbin专题和洛谷专题的刷题记录,以及简单搜索总结

知识点

  • BFS
  • DFS
  • 路径还原

白书例题(DFS + 简单剪枝)

给定整数a1、a2…、an,判断是否可以从中选出若干数,使他们的和恰好为k。

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#include<iostream>
#include<algorithm>
using namespace std;

const int MAX_N = 20;

int n;
int a[MAX_N];
int k;

bool dfs(int i, int sum)
{
if(sum > k) return false;//剪枝
if(i == n) return sum == k;
if(dfs(i+1, sum)) return true;
if(dfs(i+1, sum+a[i])) return true;
return false;
}

int main()
{
cin >> n;
for(int i=0; i<n; ++i) cin >> a[i];
cin >> k;
if(dfs(0, 0))
printf("yes\n");
else
printf("no\n");
return 0;
}

POJ1321(DFS,八皇后变式)

类似n皇后问题,区别在于某行可不放棋子

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#include<cstdio>
#include<cstring>

const int MAX_N = 8;

char map[MAX_N+1][MAX_N+1];
int n, k, ans = 0;
bool check[MAX_N+1];

void dfs(int x, int now)
{
if(x == n && now != k)
return ;
if(now == k)
{
ans++;
return ;
}
else
{
for(int i = 0; i < n; ++i)
{
if(map[x][i] == '#' && !check[i])
{
check[i] = 1;
dfs(x+1, now+1);
check[i] = 0;
}
}
dfs(x+1, now);
}
}

int main()
{
while(true)
{
scanf("%d%d", &n, &k);
if(n == -1 && k == -1)
break;
for(int i = 0; i < n; ++i)
{
scanf("%s", map[i]);
}
ans = 0;
memset(check, 0, sizeof(check));
dfs(0, 0);
printf("%d\n", ans);
}
return 0;
}

POJ2251(三维BFS)

简单扩展下二维BFS,用结构体表示坐标即可

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#include<cstdio>
#include<queue>
using namespace std;

struct Node
{
int l;
int r;
int c;
};

const int MAX_LRC = 30 + 1;
const int INF = 1e9;

char maze[MAX_LRC][MAX_LRC][MAX_LRC];
int t[MAX_LRC][MAX_LRC][MAX_LRC];
int L, R, C;
int sl, sr, sc;
int el, er, ec;

int dl[] = {1, -1, 0, 0, 0, 0};
int dr[] = {0, 0, 1, -1, 0, 0};
int dc[] = {0, 0, 0, 0, 1, -1};

void bfs()
{
queue<Node> que;
Node q;
q.l = sl;
q.r = sr;
q.c = sc;
que.push(q);
while(que.size())
{
Node qu = que.front();
que.pop();
for(int i = 0; i < 6; ++i)
{
Node qe;
qe.l = qu.l + dl[i];
qe.r = qu.r + dr[i];
qe.c = qu.c + dc[i];
if(qe.l >= 0 && qe.l < L && qe.r >= 0 && qe.r < R
&& qe.c >= 0 && qe.c < C && maze[qe.l][qe.r][qe.c] != '#'
&& t[qe.l][qe.r][qe.c] == INF)
{
que.push(qe);
t[qe.l][qe.r][qe.c] = t[qu.l][qu.r][qu.c] + 1;
}
}
}
}

int main()
{
while(true)
{
scanf("%d%d%d", &L, &R, &C);
if(L == 0)
break;
for(int i = 0; i < L; ++i)
{
for(int j = 0; j < R; ++j)
{
scanf("%s", maze[i][j]);
for(int k = 0; k < C; ++k)
{
t[i][j][k] = INF;
if(maze[i][j][k] == 'S')
{
sl = i;
sr = j;
sc = k;
}
else if(maze[i][j][k] == 'E')
{
el = i;
er = j;
ec = k;
}
}
}
}
t[sl][sr][sc] = 0;
bfs();
if(t[el][er][ec] == INF)
printf("Trapped!\n");
else
printf("Escaped in %d minute(s).\n", t[el][er][ec]);
}
return 0;
}

POJ3278(非图类BFS)

其实这题就是步长不同的BFS

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#include<cstdio>
#include<queue>
using namespace std;

const int MAX_N = 1e5 + 1;

int n, k;
int ans[MAX_N];

void bfs()
{
queue<int> que;
que.push(n);
while(que.size())
{
int p = que.front();
que.pop();
if(p == k)
break;
for(int i = 0; i < 3; ++i)
{
int t;
if(i == 0)
t = p - 1;
else if(i == 1)
t = p + 1;
else
t = p * 2;
if(t >= 0 && t <= MAX_N && !ans[t])
{
que.push(t);
ans[t] = ans[p] + 1;
}
}
}
}

int main()
{
scanf("%d%d", &n, &k);
bfs();
printf("%d\n", ans[k]);
return 0;
}

POJ3279(集合的整数表示 + 枚举)

感觉不像搜索……

需要枚举集合,用到集合的整数表示

第一行情况固定了,后面每一行的翻转情况也就确定了

我的方法比较傻,所以再贴一个白书上的题解

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#include<cstdio>
#include<cstring>

const int MAX_N = 15 + 1;

int rec[MAX_N][MAX_N];
int rect[MAX_N][MAX_N];
int t[MAX_N][MAX_N];
int ans[MAX_N][MAX_N];
int dx[] = {1, 0, 0, -1, 0};
int dy[] = {0, 1, 0, 0, -1};

int N, M;

bool check(int x, int y) //检查是否合法
{
return (0 <= x && x < M && 0 <= y && y < N);
}

void flip(int x, int y) //翻转上下左右和自己
{
for(int i = 0; i < 5; ++i)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(check(nx, ny))
{
rect[nx][ny] = !rect[nx][ny];
}
}
}

int solve()
{
for(int i = 1; i < M; ++i) //第二行开始
{
for(int j = 0; j < N; ++j)
{
if(rect[i - 1][j]) //如果上面一行的这列是黑的,则翻转
{
flip(i, j);
t[i][j] = 1;
}
}
}

for(int j = 0; j < N; ++j)
{
if(rect[M - 1][j])
return -1;
}

int res = 0;
for(int i = 0; i < M; ++i)
{
for(int j = 0; j < N; ++j)
{
if(t[i][j])
res += 1;
}
}
return res;
}

int main()
{
int res = -1;

scanf("%d%d", &M, &N);
for(int i = 0; i < M; ++i)
{
for(int j = 0; j < N; ++j)
{
scanf("%d", &rec[i][j]);
}
}

for(int i = 0; i < 1 << N; ++i) //枚举每个子集,注意,最小字典序应从最右边一位枚举
{
memset(t, 0, sizeof(t));
memcpy(rect, rec, sizeof(rec)); //建立副本
for(int j = 0; j < N; ++j) //第一行的情况
{
t[0][N - 1 - j] = i >> j & 1; //i右移j个单位,那个位上的值
}

for(int k = 0; k < N; ++k) //翻转第一行
{
if(t[0][k])
flip(0, k);
}

// for(int j = 0; j < M; ++j)
// {
// for(int k = 0; k < N; ++k)
// {
// printf("%d ", rect[j][k]);
// }
// printf("\n");
// }
// printf("\n");

int num = solve(); //根据第一行推断后面几行
if(num >= 0 && (res < 0 || res > num))
{
res = num;
memcpy(ans, t, sizeof(t));
}
}

if(res == -1)
printf("IMPOSSIBLE\n");
else
{
for(int j = 0; j < M; ++j)
{
for(int k = 0; k < N; ++k)
{
printf("%d ", ans[j][k]);
}
printf("\n");
}
}
return 0;
}
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#include<cstdio>
#include<cstring>

const int MAX_N = 15 + 1;
const int MAX_M = 15 + 1;
const int dx[5] = {-1, 0, 0, 0, 1};
const int dy[5] = {0, -1, 0, 1, 0};

int M, N;
int tile[MAX_M][MAX_N];

int opt[MAX_M][MAX_N];
int flip[MAX_M][MAX_N];

int get(int x, int y) //查询(x, y)颜色
{
int c = tile[x][y];
for(int d = 0; d < 5; ++d)
{
int x2 = x + dx[d];
int y2 = y + dy[d];
if(0 <= x2 && x2 < M && 0 <= y2 && y2 < N)
{
c += flip[x2][y2];
}
}
return c % 2;
}

int calc() //求出第二行开始的翻转方法
{
for(int i = 1; i < M; ++i) //若为黑,则翻转
{
for(int j = 0; j < N; ++j)
{
if(get(i - 1, j) != 0)
{
flip[i][j] = 1;
}
}
}

for(int j = 0; j < N; ++j) //判断最后一行是否全白
{
if(get(M - 1, j) != 0)
return -1;
}

int res = 0;
for(int i = 0; i < M; ++i) //统计次数
{
for(int j = 0; j < N; ++j)
{
res += flip[i][j];
}
}
return res;
}

void solve()
{
int res = -1;

for(int i = 0; i < 1 << N; ++i)
{
memset(flip, 0, sizeof(flip));
for(int j = 0; j < N; ++j)
{
flip[0][N - j - 1] = i >> j & 1;
}

int num = calc();
if(num >= 0 && (res < 0 || res > num))
{
res = num;
memcpy(opt, flip, sizeof(flip));
}
}

if(res < 0)
{
printf("IMPOSSIBLE\n");
}
else
{
for(int i = 0; i < M; ++i)
{
for(int j = 0; j < N; ++j)
{
printf("%d%c", opt[i][j], j + 1 == N ? '\n' : ' ');
}
}
}
}

int main()
{
scanf("%d%d", &M, &N);
for(int i = 0; i < M; ++i)
{
for(int j = 0; j < N; ++j)
{
scanf("%d", &tile[i][j]);
}
}
solve();
return 0;
}

POJ1426(BFS + 思考)

再字符串后分别加0和1,然后判断余数,若余数没有出现过,则是一个新状态,加入队列

可以用除法列竖式来理解算法思路

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#include<cstdio>
#include<iostream>
#include<string>
#include<queue>
#include<cstring>
using namespace std;

int n;
int book[300];

struct node
{
int y;
string ans;
node(int y, string ans) : y(y), ans(ans){}
};

void solve()
{
queue<node> que;
que.push(node(1, "1"));
int yy; //新余数

while(que.size())
{
node t = que.front();
que.pop();

if(!t.y) //如果余数是0,则输出
{
cout << t.ans << '\n';
return ;
}

yy = (t.y * 10 + 1) % n; //末尾加1的情况
if(!book[yy])
{
book[yy] = 1;
que.push(node(yy, t.ans + "1")); //记录路径
}

yy = (t.y * 10) % n; //末尾加0的情况
if(!book[yy])
{
book[yy] = 1;
que.push(node(yy, t.ans + "0"));
}
}
}

int main()
{
while(true)
{
scanf("%d", &n);
if(!n) break;
memset(book, 0, sizeof(book));
solve();
}
return 0;
}

POJ3126(素数筛 + 非图类BFS)

枚举改变一个数字后的所有状态,判断是否是素数,是素数则放入队列

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#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;

const int MAX_N = 10000;

bool prime[MAX_N];
int d[MAX_N];

void init()
{
for(int i = 1; i <= 9999; ++i)
{
prime[i] = true;
}
prime[1] = false;
}

void prim()
{
init();
for(int i = 2; i <= 9999; ++i)
{
if(prime[i])
{
for(int j = 2; j * i <= 9999; ++j)
{
prime[j * i] = false;
}
}
}
}

int n;
int a, b;

int solve(int a, int b)
{
if(a == b)
return 0;
queue<int> que;
que.push(a);
d[a] = 0;
while(que.size())
{
int t = que.front();
que.pop();
int temp = t / 1000;
for(int i = 1; i < 10; ++i)
{
if(i == temp)
continue;
int t1 = i * 1000 + t - temp * 1000;
if(prime[t1] && d[t1] == -1)
{
d[t1] = d[t] + 1;
que.push(t1);
}
}
temp = t % 1000 / 100;
for(int i = 0; i < 10; ++i)
{
if(i == temp)
continue;
int t1 = i * 100 + t - temp * 100;
if(prime[t1] && d[t1] == -1)
{
d[t1] = d[t] + 1;
que.push(t1);
}
}
temp = t % 100 / 10;
for(int i = 0; i < 10; ++i)
{
if(i == temp)
continue;
int t1 = i * 10 + t - temp * 10;
if(prime[t1] && d[t1] == -1)
{
d[t1] = d[t] + 1;
que.push(t1);
}
}
temp = t % 10;
for(int i = 0; i < 10; ++i)
{
if(i == temp )
continue;
int t1 = t - temp + i;
if(prime[t1] && d[t1] == -1)
{
d[t1] = d[t] + 1;
que.push(t1);
}
}
}
return d[b];
}

int main()
{
prim(); //筛素数
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%d%d", &a, &b);
fill(d, d + MAX_N, -1);
int ans = solve(a, b);
if(ans == -1)
printf("Impossible\n");
else
printf("%d\n", ans);
}
return 0;
}

POJ3087(打表 + 模拟?)

我已经想到一个绝妙的证明,可惜这里空间太小写不下(皮)

是道模拟题吧,不知道为什么归在搜索。编写很简单,但是证明了一个多小时还证不出来,题解也没有相关证明。

设a1,a2,…a2n

则新串b1,b2,…b2n

有bi =a (i & 1 ? i = (i + 1) / 2 : i / 2);

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#include<cstdio>

int main()
{
int n;
n = 10;
int i = 2;
while(true)
{
if(i & 1)
i = (i + 1) / 2 + n; //新i
else
i = i / 2; //新i
printf("%d\n", i);
}
return 0;
}

打表发现总有一个新i会等于旧i,也就是说,总会有bi = … = ai

用反证法。设i0,i的范围为(0,2n],如果i0经过2n次变换,ik始终不等于i0,那么必定有个i出现了两次,而这个i一定由i0变化而来,所以i0 在变化中出现了循环,也即在2n次变换中,一定存在bi = … = ai

但是思路有点问题,bi = … = ai所用的次数各不相同,无法证明所有的b和a相等……

看看学了离散和具体数学后有没有办法证明吧。

AC代码

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//不严谨,待证明

#include<iostream>
#include<string>
using namespace std;

string str[1000];
int t;
int len;
string str1, str2, str3;

bool find(string strn, int sum)
{
for(int i = 0; i < sum; ++i)
{
if(str[i] == strn)
return true;
}
return false;
}

int main()
{
cin >> t;
for(int i = 0; i < t; ++i)
{
cin >> len;
cin >> str1 >> str2 >> str3;
int sum = 0; //新字串的数量
while(true)
{
string strt = str3;
for(int j = 0; j < len; ++j)
{
strt[j*2] = str2[j];
strt[j*2+1] = str1[j];
}
str[sum] = strt;
if(str[sum] == str3)
{
cout << i+1 << ' ' << sum+1 << '\n';
break;
}
else if(find(str[sum], sum))
{
cout << i+1 << ' ' << -1 << '\n';
break;
}
sum++;
for(int j = 0; j < len; ++j)
{
str1[j] = strt[j];
str2[j] = strt[len + j];
}
}
}
return 0;
}

POJ3414(BFS + 思考)

参考自博客https://blog.csdn.net/qq_41280600/article/details/102491233

新思路,变量结构体中一个值储存路径

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//新思路,bfs中,用变量记录每条路径
//https://blog.csdn.net/qq_41280600/article/details/102491233
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<queue>
using namespace std;

struct node //初始化顺序要对应,不然会warning
{
int a, b;
string opt;
int step;
node(int a, int b, string opt, int step): a(a), b(b), opt(opt), step(step){}
};

int a, b, c;
int vis[105][105];

string bfs()
{
queue<node> que;
que.push(node(0, 0, "", 0));
vis[0][0] = 1;

while(que.size())
{
node t = que.front();
que.pop();
if(t.a == c || t.b == c)
{
cout << t.step <<endl; //输出步数
return t.opt; //返回路径
}

/* Fill */
if(!vis[a][t.b])
{
vis[a][t.b] = 1;
que.push(node(a, t.b, t.opt + "FILL(1)\n", t.step + 1));
}
if(!vis[t.a][b])
{
vis[t.a][b] = 1;
que.push(node(t.a, b, t.opt + "FILL(2)\n", t.step + 1));
}

/* DROP */
if(t.a != 0 && !vis[0][t.b])
{
vis[0][t.b] = 1;
que.push(node(0, t.b, t.opt + "DROP(1)\n", t.step + 1));
}
if(t.b != 0 && !vis[t.a][0])
{
vis[t.a][0] = 1;
que.push(node(t.a, 0, t.opt + "DROP(2)\n", t.step + 1));
}

/* POUR */
if(t.a != 0 && t.b != b)
{
if(t.a + t.b <= b)
{
if(!vis[0][t.a + t.b])
{
vis[0][t.a + t.b] = 1;
que.push(node(0, t.a + t.b, t.opt + "POUR(1,2)\n", t.step + 1));
}
}
else
{
if(!vis[t.a + t.b - b][b])
{
vis[t.a + t.b - b][b] = 1;
que.push(node(t.a + t.b - b, b, t.opt + "POUR(1,2)\n", t.step + 1));
}
}
}
if(t.b != 0 && t.a != a)
{
if(t.a + t.b <= a)
{
if(!vis[t.a + t.b][0])
{
vis[t.a + t.b][0] = 1;
que.push(node(t.a + t.b, 0, t.opt + "POUR(2,1)\n", t.step + 1));
}
}
else
{
if(!vis[a][t.a + t.b - a])
{
vis[a][t.a + t.b - a] = 1;
que.push(node(a, t.a + t.b - a, t.opt + "POUR(2,1)\n", t.step + 1));
}
}
}
}
return "impossible";
}

int main()
{
scanf("%d%d%d", &a, &b, &c);
cout << bfs();
return 0;
}

FZU2150(双点BFS)

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//连通集大于2个则输出-1
//连通集等于两个则各自bfs
//连通集等于1个则?爆搜!
//双起点bfs
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;

const int MAX_N = 20;
const int INF = 2e9 + 7;

int T, n, m;
char maze[MAX_N][MAX_N];
char book[MAX_N][MAX_N];
int ans;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
int d[MAX_N][MAX_N];

typedef pair<int, int> pr;

queue<pr> que;

void my_fill(int x, int y)
{
book[x][y] = '.';
for(int i = 0; i < 4; ++i)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(0 <= nx && nx < n && 0 <= ny && ny < m
&& book[nx][ny] == '#')
my_fill(nx, ny);
}
}

bool check() //检查连通集是否大于两个
{
int count = 0;
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < m; ++j)
{
if(book[i][j] == '#')
{
my_fill(i, j);
count++;
}
}
}
if(count > 2) return false;
return true;
}

int solve(int x, int y, int a, int b)
{
int ret = 0;
queue<pr> p;
p.push(pr(x, y));
p.push(pr(a, b));
d[x][y] = d[a][b] = 0;

while(p.size())
{
pr t = p.front();
p.pop();
for(int i = 0; i < 4; ++i)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if(0 <= nx && nx < n && 0 <= ny && ny < m && maze[nx][ny] == '#'
&& d[nx][ny] == -1)
{
ret = d[nx][ny] = d[t.first][t.second] + 1;
p.push(pr(nx, ny));
}
}
}

for(int i = 0; i < n; ++i) //检查是否有点没有搜过(两个起点都在一个连通集的情况)
{
for(int j = 0; j < m; ++j)
{
if(maze[i][j] == '#' && d[i][j] == -1)
{
return INF;
}
}
}
return ret;
}

int main()
{
scanf("%d", &T);
for(int i = 0; i < T; ++i)
{
que = queue<pr>(); //初始化
ans = INF;

scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
{
scanf("%s", maze[i]);
for(int j = 0; j < m; ++j)
{
book[i][j] = maze[i][j];
if(maze[i][j] == '#')
{
que.push(pr(i, j));
}
}
}

if(!check())
{
printf("Case %d: -1\n", i + 1);
continue;
}

if(que.size() <= 2)
{
printf("Case %d: 0\n", i + 1);
continue;
}

while(que.size()) //总共que.size()个草的坐标,列举所有组合
{
pr t1 = que.front();
que.pop();
int count = que.size();
while(count--)
{
pr t2 = que.front();
que.pop();
que.push(t2);
memset(d, -1, sizeof(d));
ans = min(ans, solve(t1.first, t1.second, t2.first, t2.second));
}
}
printf("Case %d: %d\n", i + 1, ans);
}
return 0;
}

UVA11624(多点BFS)

F的初始位置有很多,用一个队列来存储F,j可达到的地方也用一个队列表示

然后就是双起点BFS了

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#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

const int MAX_N = 1000 + 1;

typedef pair<int, int> pr;

int n;
int r, c;

char maze[MAX_N][MAX_N];
int d[MAX_N][MAX_N];
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
int jx, jy;
int ret = 0;
queue<pr> F;

bool solve()
{
queue<pr> qj;
qj.push(pr(jx, jy));
d[jx][jy] = 0;

while(qj.size())
{
int count = F.size();
while(count--)
{
pr t2 = F.front();
F.pop();
for(int i = 0; i < 4; ++i)
{
int nx = t2.first + dx[i];
int ny = t2.second + dy[i];
if(0 <= nx && nx < r && 0 <= ny && ny < c
&& (maze[nx][ny] == '.' || maze[nx][ny] == 'J')) //火势可蔓延
{
maze[nx][ny] = 'F';
F.push(pr(nx, ny));
}
}
}

count = qj.size();
while(count--)
{
pr t = qj.front();
qj.pop();
for(int i = 0; i < 4; ++i)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if(0 <= nx && nx < r && 0 <= ny && ny < c)
{
if(d[nx][ny] == -1 && maze[nx][ny] == '.')
{
d[nx][ny] = d[t.first][t.second] + 1;
qj.push(pr(nx, ny));
}
}
else
{
ret = d[t.first][t.second] + 1;
return true;
}
}
}
}
return false;
}

int main()
{
// freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
F = queue<pr>();
memset(d, -1, sizeof(d));

scanf("%d%d", &r, &c);
for(int j = 0; j < r; ++j)
{
scanf("%s", maze[j]);
for(int k = 0; k < c; ++k)
{
if(maze[j][k] == 'F')
{
F.push(pr(j, k));
}
if(maze[j][k] == 'J')
{
jx = j;
jy = k;
}
}
}
if(solve())
printf("%d\n", ret);
else
printf("IMPOSSIBLE\n");
}
return 0;
}

POJ3984(DFS + 路径还原)

路径还原

我选择用一个栈同步DFS……

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#include<cstdio>
#include<stack>
using namespace std;

const int MAX_N = 6;

int maze[MAX_N][MAX_N];
int d[MAX_N][MAX_N];

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

int mini = 2e9;

typedef pair<int, int> pr;
stack<pr> s1, s2;

void dfs(int x, int y)
{
//printf("(%d, %d)\n", x, y);
if(x == 4 && y == 4)
{
if((int)s1.size() < mini)
{
s2 = s1;
mini = s1.size();
}
return ;
}
else
{
for(int i = 0; i < 4; ++i)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(0 <= nx && nx < 5 && 0 <= ny && ny < 5
&& !maze[nx][ny] && !d[nx][ny])
{
s1.push(pr(nx, ny));
d[nx][ny] = 1;
dfs(nx, ny);
d[nx][ny] = 0;
s1.pop();
}
}
}
}

int main()
{
for(int i = 0; i < 5; ++i)
for(int j = 0; j < 5; ++j)
scanf("%d", &maze[i][j]);
s1.push(pr(0, 0));
d[0][0] = 1;
dfs(0, 0);

stack<pr> s3;
while(!s2.empty())
{
s3.push(s2.top());
s2.pop();
}
while(!s3.empty())
{
printf("(%d, %d)\n", s3.top().first, s3.top().second);
s3.pop();
}
return 0;
}

HDU1241(DFS求连通块)

简单的连通块搜索

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#include<cstdio>

const int MAX_N = 100 + 1;

int n, m;
char map[MAX_N][MAX_N];

void dfs(int x, int y)
{
for(int i = -1; i <= 1; ++i)
{
for(int j = -1; j <= 1; ++j)
{
int nx = x + i;
int ny = y + j;
if(0 <= nx && nx < n && 0 <= ny && ny < m
&& map[nx][ny] == '@')
{
map[nx][ny] = '*';
dfs(nx, ny);
}
}
}
}

int main()
{
while(true)
{
int ans = 0;
scanf("%d%d", &n, &m);
if(!n) break;
for(int i = 0; i < n; ++i)
{
scanf("%s", map[i]);
}
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < m; ++j)
{
if(map[i][j] == '@')
{
dfs(i, j);
ans++;
}
}
}
printf("%d\n", ans);
}
return 0;
}

HDU1495(BFS + 枚举 + 思考)

枚举倒完水后各个容器内水的体积,然后BFS即可

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#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;

struct node
{
int x;
int y;
int z;
};

int ans[102][102][102];

int solve(int S, int N, int M)
{
if(S & 1) return -1;

memset(ans, -1, sizeof(ans));
ans[S][0][0] = 0;

queue<node> que;
node t;
t.x = S, t.y = 0, t.z = 0;
que.push(t);
while(que.size())
{
t = que.front();
que.pop();
int count = 0;
if(t.x == S / 2)
count++;
if(t.y == S / 2)
count++;
if(t.z == S / 2)
count++;
if(count == 2)
return ans[t.x][t.y][t.z];

node t1;
if(t.x + t.y <= N) //A倒B
{
t1.x = 0;
t1.y = t.x + t.y;
t1.z = t.z;
}
else
{
t1.x = t.x + t.y - N;
t1.y = N;
t1.z = t.z;
}
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}

if(t.x + t.z <= M) //A倒C
{
t1.x = 0;
t1.y = t.y;
t1.z = t.x + t.z;
}
else
{
t1.x = t.x + t.z - M;
t1.y = t.y;
t1.z = M;
}
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}

if(t.y + t.z <= M) //B倒C
{
t1.x = t.x;
t1.y = 0;
t1.z = t.y + t.z;
}
else
{
t1.x = t.x;
t1.y = t.y + t.z - M;
t1.z = M;
}
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}

if(t.y + t.z <= N) //C倒B
{
t1.x = t.x;
t1.y = t.y + t.z;
t1.z = 0;
}
else
{
t1.x = t.x;
t1.y = N;
t1.z = t.y + t.z - N;
}
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}

t1.x = t.x + t.y; //B倒A
t1.y = 0;
t1.z = t.z;
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}

t1.x = t.x + t.z; //C倒A
t1.y = t.y;
t1.z = 0;
if(ans[t1.x][t1.y][t1.z] == -1)
{
que.push(t1);
ans[t1.x][t1.y][t1.z] = ans[t.x][t.y][t.z] + 1;
}
}
return -1;
}

int main()
{
int S, N, M;
while(true)
{
scanf("%d%d%d", &S, &N, &M);
if(!S)
break;
int ret = solve(S, N, M);
if(ret == -1)
printf("NO\n");
else
printf("%d\n", ret);
}
return 0;
}

HDU2612(两点BFS)

可能有的@点无法到达,要排除

两个BFS,把结果相加即可

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//简单bfs
//易错!!注意可能有@无法到达,要特判
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;

const int MAX_N = 200 + 10;

int n, m;
char maze[MAX_N][MAX_N];
int ansY[MAX_N][MAX_N];
int ansM[MAX_N][MAX_N];
int YX, YY;
int MX, MY;
int ret = 1e9;

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

typedef pair<int, int> pr;

void solve(int x, int y, int (& ans)[MAX_N][MAX_N])
{
queue<pr> que;
que.push(pr(x, y));
while(que.size())
{
pr t = que.front();
que.pop();
for(int i = 0; i < 4; ++i)
{
int nx = t.first + dx[i];
int ny = t.second + dy[i];
if(0 <= nx && 0 <= ny && nx < n && ny < m && maze[nx][ny] != '#'
&& maze[nx][ny] != 'Y' &&maze[nx][ny] != 'M' && ans[nx][ny] == -1)
{
ans[nx][ny] = ans[t.first][t.second] + 1;
que.push(pr(nx, ny));
}
}
}
}

int main()
{
while(scanf("%d%d", &n, &m) != EOF)
{
for(int i = 0; i < n; ++i)
{
scanf("%s", maze[i]);
for(int j = 0; j < m; ++j)
{
if(maze[i][j] == 'Y')
{
YX = i;
YY = j;
}
if(maze[i][j] == 'M')
{
MX = i;
MY = j;
}
}
}
memset(ansY, -1, sizeof(ansY));
ansY[YX][YY] = 0;
solve(YX, YY, ansY);
memset(ansM, -1, sizeof(ansM));
ansM[MX][MY] = 0;
solve(MX, MY, ansM);
ret = 1e9;
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < m; ++j)
{
if(maze[i][j] == '@' && ansY[i][j] != -1 && ansM[i][j] != -1)
ret = min(ret, ansY[i][j] + ansM[i][j]);
}
}
printf("%d\n", ret * 11);
}
return 0;
}

洛谷P1219(DFS,经典八皇后问题)

经典n皇后问题

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#include<cstdio>

int n, sum;
int ans[14];
bool check[3][30];

void solve(int line)
{
if(line > n)
{
sum++;
if(sum > 3)
return ;
else
{
for(int i = 1; i <= n; i++)
{
printf("%d ", ans[i]);
}
printf("\n");
return ;
}
}
for(int i = 1; i <= n; i++)
{
if(!check[0][i] && !check[1][line+i] && !check[2][line+n-i])
{
ans[line] = i;
check[0][i] = check[1][line+i] = check[2][line+n-i] = 1;
solve(line+1);
check[0][i] = check[1][line+i] = check[2][line+n-i] = 0;
}
}
}

int main()
{
scanf("%d", &n);
solve(1);
printf("%d\n", sum);
return 0;
}

洛谷P1019(DFS,暴力)

注意linkstr()函数返回的应该是最小重叠部分,而非最大

注意开头的字符串也要标记使用过一次

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#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

const int MAX_N = 20 + 1;

int n;
int ans = 0;
char ch;
string str[MAX_N];
int used[MAX_N];

int linkstr(string str1, string str2)
{
for(int i = 0; i < min((int)str1.length(), (int)str2.length()); ++i)
{
int flag = 1;
for(int j = i; j >= 0; --j)
{
if(str2[i-j] != str1[str1.length() - j - 1])
flag = 0;
}
if(flag)
return i + 1;
}
}

void dfs(string strnow, int lengthnow)
{
ans = max(lengthnow, ans);
for(int i = 0; i < n; ++i)
{
int len = linkstr(strnow, str[i]);
if(used[i] && len)
{
used[i]--;
dfs(str[i], str[i].length() + lengthnow - len);
used[i]++;
}
}
}

int main()
{
cin >> n;
for(int i = 0; i < n; ++i)
{
cin >> str[i];
used[i] = 2;
}
cin >> ch;
for(int i = 0; i < n; ++i)
{
if(str[i][0] == ch)
{
used[i]--;
dfs(str[i], str[i].length());
used[i]++;
}
}
cout << ans << endl;
return 0;
}
//最小重叠部分
//第一个开头的也要计算使用过

洛谷P1101(单向DFS)

使用字符串常量

参数包含方向

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#include<cstdio>

const int MAX_N = 100 + 1;

char map[MAX_N][MAX_N];
char str[] = "yizhong";
char ans[MAX_N][MAX_N];
int n;

void dfs(int x, int y, int now, int dx, int dy)
{
if(now == 6)
{
int nx = x;
int ny = y;
ans[nx][ny] = 'g';
for(int i = 1; i <= 6; ++i)
{
nx -= dx;
ny -= dy;
ans[nx][ny] = str[6-i];
}
}
else
{
int nx = x + dx;
int ny = y + dy;
if(nx >= 0 && nx < n && ny >= 0 && ny < n
&& map[nx][ny] == str[now + 1])
dfs(nx, ny, now + 1, dx, dy);
}
}

int main()
{
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%s", map[i]);
for(int j = 0; j < n; ++j)
{
ans[i][j] = '*';
}
}
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(map[i][j] == 'y')
{
for(int k = -1; k <= 1; ++k)
{
for(int l = -1; l <= 1; ++l)
{
if(!k && !l)
continue;
dfs(i, j, 0, k, l);
}
}
}
}
}
for(int i = 0; i < n; ++i)
{
puts(ans[i]);
}
return 0;
}

洛谷P1605(DFS求路径数)

简单题

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#include<cstdio>

const int MAX_NM = 5 + 2;
const int INF = 1e9;

int map[MAX_NM][MAX_NM];
int d[MAX_NM][MAX_NM];
int n, m, t;
int sx, sy, gx, gy;
int ans = 0;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

void dfs(int x, int y)
{
if(x == gx && y == gy)
{
ans++;
return ;
}
else
{
for(int i = 0; i < 4; ++i)
{
int nx = x + dx[i];
int ny = y + dy[i];
if(nx > 0 && nx <= n && ny > 0 && ny <= m && map[nx][ny] != INF
&& !d[nx][ny])
{
d[nx][ny] = 1;
dfs(nx, ny);
d[nx][ny] = 0;
}
}
}
}

int main()
{
scanf("%d%d%d", &n, &m, &t);
scanf("%d%d%d%d", &sx, &sy, &gx, &gy);
for(int i = 0; i < t; ++i)
{
int x, y;
scanf("%d%d", &x, &y);
map[x][y] = INF;
}
d[sx][sy] = 1;
dfs(sx, sy);
printf("%d\n", ans);
return 0;
}
//起点坐标不要忘记标记使用过

洛谷P1162(BFS求闭合圈)

先把边界预先染成其它颜色

染色时,先染色再push(),避免重复计算

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#include<cstdio>
#include<queue>
using namespace std;

const int MAX_N = 30 + 1;

typedef pair<int, int> pr;

int n;
int map[MAX_N][MAX_N];
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};

void bfs(int x, int y, int c)
{
queue<pr> que;
que.push(pr(x, y));
while(que.size())
{
pr p = que.front();
que.pop();
map[p.first][p.second] = c;
for(int i = 0; i < 4; ++i)
{
int nx = p.first + dx[i];
int ny = p.second + dy[i];
if(0 <= nx && nx < n && 0 <= ny && ny < n
&& map[nx][ny] == 0)
{
map[nx][ny] = c;
que.push(pr(nx, ny));
}
}
}
}

int main()
{
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
scanf("%d", &map[i][j]);
}
}
for(int i = 0; i < n; ++i)
{
if(!map[i][0])
bfs(i, 0, 3);
if(!map[i][n-1])
bfs(i, n-1, 3);
if(!map[0][i])
bfs(0, i, 3);
if(!map[n-1][i])
bfs(n-1, i, 3);
}
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(map[i][j] == 0)
{
bfs(i, j, 2);
}
}
}
for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(map[i][j] == 3)
printf("0 ");
else
printf("%d ", map[i][j]);
}
printf("\n");
}
return 0;
}
//先染色再放进去,不然会超时甚至爆栈

洛谷P1141(BFS,染色)

读错题了orz,不是求从(x, y)所能达到的最远距离,而是从(x, y)所能达到的最多格子数

那么求连通集就好,一个连通集内,所能达到的地方相同

比如在学校是一个连通集,不管是从教学楼出发,还是从寝室出发,最终能到达的地方都是整个学校

  • %1d这样的输入格式
  • 小技巧,给d赋不同的值,然后映射ret
  • ret数组不要开小了……
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#include<cstdio>
#include<queue>
using namespace std;

const int MAX_N = 1e3 + 1;

int maze[MAX_N][MAX_N];
int d[MAX_N][MAX_N];
int ret[MAX_N * MAX_N + 1];

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

int ans = 1;
int n, m;
int num = 1;

typedef pair<int, int> pr;

queue<pr> que;

void bfs(int x, int y, int num)
{
d[x][y] = num;
que.push(pr(x, y));
while(!que.empty())
{
pr t = que.front();
que.pop();
int nx, ny;
for(int i = 0; i < 4; ++i)
{
nx = t.first + dx[i];
ny = t.second + dy[i];
if(0 <= nx && nx < n && 0 <= ny && ny < n && !d[nx][ny]
&& maze[nx][ny] != maze[t.first][t.second])
{
d[nx][ny] = num;
ans++;
que.push(pr(nx, ny));
}
}
}
ret[num] = ans;
ans = 1;
}

int main()
{
scanf("%d%d", &n, &m);
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
scanf("%1d", &maze[i][j]); //输入格式

for(int i = 0; i < n; ++i)
{
for(int j = 0; j < n; ++j)
{
if(!d[i][j])
bfs(i, j, num++);
}
}

for(int i = 0; i < m; ++i)
{
int a, b;
scanf("%d%d", &a, &b);
printf("%d\n", ret[d[a-1][b-1]]);
}
return 0;
}

洛谷P1443(非上下左右移动的BFS)

改变方向数据即可

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#include<cstdio>
#include<queue>
#include<utility>
using namespace std;

const int MAX_MN = 400 + 1;

typedef pair<int, int> pr;

int map[MAX_MN][MAX_MN];
int m, n, sx, sy;
int dx[] = {2, 2, 1, 1, -1, -1, -2, -2};
int dy[] = {1, -1, 2, -2, 2, -2, 1, -1};

void bfs()
{
queue<pr> que;
que.push(pr(sx-1, sy-1));
while(que.size())
{
pr p = que.front();
que.pop();
for(int i = 0; i < 8; ++i)
{
int nx = p.first + dx[i];
int ny = p.second + dy[i];
if(0 <= nx && nx < m && 0 <= ny && ny < n
&& map[nx][ny] == -1)
{
map[nx][ny] = map[p.first][p.second] + 1;
que.push(pr(nx, ny));
}
}
}
}

int main()
{
scanf("%d%d%d%d", &m, &n, &sx, &sy);
for(int i = 0; i < m; ++i)
{
for(int j = 0; j < n; ++j)
{
map[i][j] = -1;
}
}
map[sx-1][sy-1] = 0;
bfs();
for(int i = 0; i < m; ++i)
{
for(int j = 0; j < n; ++j)
{
printf("%-5d", map[i][j]);
}
printf("\n");
}
return 0;
}
//注意格式

总结

何为搜索?

我接触的第一道搜索题是白书上的部分和问题。是一道简单的BFS,当时由于对递归的理解不够,导致花了好多天才理解为什么要这么写。到我能真正写出DFS那是更后面的事了。

后来我会熟练地写地图上的搜索了,并天真地以为搜索就是指图的搜索。在一个图中给出入口出口,寻找最短路径或者求连通块便是搜索。我做了很多板子题,并自得其乐。洛谷P1101洛谷P1605洛谷P1162洛谷P1141

再后来逐渐接触一些稍微变形点的题,比如三维的啦POJ2251,像马那样跳跃的啦洛谷P1443,多点一起的搜索啦FZU2150UVA11624,八皇后相关的啦洛谷P1219POJ1321,路径还原啦POJ3984POJ3414

再后来,我就自闭了……因为发现这个时候剩下的题都是没什么思路的,它不是一个地图,我想象不出它的样子,无法把它们套进搜索的模型里。

好在随着时间的流逝,我也逐渐开窍了。懂得了地图只是搜索的一个很直观的表现。搜索是一个状态根据一个规则到达另一份状态,后者再根据同一个规则到达下一份状态,直至搜完所有的可能或者得到目标值,应该也是一种暴力的方法(莽夫的智慧)。而BFS还是DFS只是搜索策略的不同罢了。