五味子

小舟从此逝,江海寄余生

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基础数据结构刷题记录

  • 队列
  • 链表
  • 并查集

洛谷P1449(栈和逆序表达式)

水题

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

int main()
{
char str[10001];
scanf("%s", str);
int len = strlen(str);
int num = 0;
stack<int> s;
for(int i = 0; i < len; ++i)
{
if(str[i] >= '0' && str[i] <= '9')
{
num *= 10;
num += str[i] - '0';
}
else if(str[i] == '.')
{
s.push(num);
num = 0;
}
else if(str[i] == '@')
{
printf("%d\n", s.top());
s.pop();
}
else
{
int a = s.top();
s.pop();
int b = s.top();
s.pop();
if(str[i] == '-')
s.push(b - a);
else if(str[i] == '+')
s.push(b + a);
else if(str[i] == '*')
s.push(b * a);
else if(str[i] == '/')
s.push(b / a);
}
}
return 0;
}

洛谷P1739(栈)

水题

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

int main()
{
int flag = 1;
stack<char> s;
char str[300];
scanf("%s", str);
int len = strlen(str);
for(int i = 0; i < len; ++i)
{
if(str[i] == '(')
s.push(str[i]);
if(str[i] == ')')
{
if(s.empty())
flag = 0;
else
s.pop();
}
}
if(!s.empty())
flag = 0;
if(flag)
printf("YES\n");
else
printf("NO\n");
return 0;
}

洛谷P1115(?)

分在洛谷的线性数据结构里,但我好像不是这么做的

  • ai表示第i个数,a[i]表示前i项和
  • ai……aj和最大的话,那么这段长度为a[j]-a[i]
  • 即a[i]最小的时候,以aj结尾的最长序列和可以一步算出
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#include<cstdio>

const int MAX_N = 2e5 + 1;
const int INF = 1e9;

int a[MAX_N];

int main()
{
int n;
scanf("%d", &n);
for(int i = 0; i < n; ++i)
{
scanf("%d", &a[i]);
a[i] = a[i] + a[i-1]; //储存前i项和
}
int min = 0;
int ans = a[0];
for(int i = 1; i < n; ++i)
{
min = min > a[i-1] ? a[i-1] : min;
ans = ans < a[i] - min ? a[i] - min : ans;
}
printf("%d\n", ans);
return 0;
}

洛谷P1160(双向链表)

简单题,初学可练手

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#include<cstdio>
#include<cstdlib>

typedef struct Node
{
int data;
Node *left;
Node *right;
} Node;

Node *book[100001]; //记录每个点的地址

Node *head = (Node *)malloc(sizeof(Node)); //头结点
Node *q = (Node *)malloc(sizeof(Node)); //第一个结点

Node *find(int num) //寻找结点
{
if(book[num])
return book[num];
return NULL;
}

void insert(int num1, int d, int num2) //插入结点
{
//在num1的左/右插入num2
Node *p = find(num1);
Node *t = (Node *)malloc(sizeof(Node));
t->data = num2;
if(!d && p->left == head)
{
t->left = head;
t->right = p;
head->right = t;
p->left = t;
}
else if(!d)
{
t->left = p->left;
t->right = p;
p->left->right = t;
p->left = t;
}
else if(d && p->right == NULL)
{
t->right = NULL;
t->left = p;
p->right = t;
}
else if(d)
{
t->right = p->right;
t->left = p;
p->right->left = t;
p->right = t;
}
book[num2] = t;
}

void del(int num) //删除结点
{
Node *p = find(num);
if(!p)
return;
if(p->left ==NULL && p->right == NULL)
{
;
}
else if(p->left == NULL)
{
p->right->left = NULL;
}
else if(p->right == NULL)
{
p->left->right = NULL;
}
else
{
p->left->right = p->right;
p->right->left = p->left;
}
free(p);
book[num] = NULL;
}

int main()
{
for(int i = 0; i < 100001; ++i)
book[i] = NULL;

head->left = NULL;
head->right = q;
q->data = 1;
q->left = head;
q->right = NULL;
book[1] = q;

int n;
scanf("%d", &n);
for(int i = 2; i < n + 1; ++i)
{
int num1, num2, d;
num2 = i;
scanf("%d%d", &num1, &d);
insert(num1, d, num2);
}

int m;
scanf("%d", &m);
for(int i = 0; i < m; ++i)
{
int num;
scanf("%d", &num);
del(num);
}

Node *p = head->right;
while(p)
{
printf("%d ", p->data);
p = p->right;
}

printf("\n");
return 0;
}

洛谷P1996(单向链表)

初学练手的简单题

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// 链表模拟解法
// 其余还有队列模拟、数组模拟解法等
#include<cstdio>
#include<cstdlib>
using namespace std;

typedef struct Node
{
int data;
struct Node * next;
} Node;

int n, m;
int size = 1;

Node *init()
{
Node *q, *head;
head = (Node *)malloc(sizeof(Node));
head->data = 1;
q = head;
for(int i = 2; i <= n; ++i)
{
Node *p = (Node *)malloc(sizeof(Node));
p->data = i;
q->next = p;
if(i == n)
p->next = head;
else
q = p;
size++;
}
return head;
}

void solve(Node *head)
{
Node *q;
Node *p = (Node *)malloc(sizeof(Node));
p = head;
while(size > 0)
{
for(int i = 1; i < m - 1; ++i)
{
q = p->next;
p = q;
}
Node *t = q->next;
q->next = q->next->next;
printf("%d ", t->data);
free(t);
size--;
if(size)
p = q->next;
}
}

int main()
{
scanf("%d%d", &n, &m);
if(!n)
return 0;
else
{
Node *head = init();
solve(head);
}
return 0;
}

洛谷P2661(图论+并查集,最小环问题)

简单题,但是要思考一下

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//图论最小环问题, 并查集
#include<cstdio>
#include<algorithm>
using namespace std;

const int MAX_N = 2e5 + 2;

int par[MAX_N];
int d[MAX_N];//记录到根节点的长度

int n;
int ans = 2e9;

void init() //初始化
{
for(int i = 0; i <= n; ++i)
{
par[i] = i;
d[i] = 0;
}
}

int find(int x) //查询根节点并更新距离
{
if(x == par[x])
return x;
else
{
int last= par[x];
par[x] = find(par[x]); //查找途中,把每个点都连向根节点
d[x] += d[last];
}
return par[x];
}

bool same(int x, int y)
{
return find(x) == find(y);
}

void unite(int x, int y)
{
x = find(x);
y = find(y);
if(x == y) return ;
else
par[x] = y; //x连到y上
}

void solve(int a, int b)
{
int x = find(a);
int y = find(b);
if(x == y)
ans = min(ans, d[a] + d[b] + 1); //环的长度为a到根节点的长度加b到根节点的长度加ab之间的1
else
{
unite(x, y);
d[a] = d[b] + 1;
}
}

int main()
{
scanf("%d", &n);
init();
for(int i = 1; i <= n; ++i)
{
int t;
scanf("%d", &t);
solve(i, t);
}
printf("%d\n", ans);
return 0;
}

洛谷P1111(并查集求连通块数量)

简单题, 想到套模板即可

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//连通块数量 = n - 1
#include<cstdio>
#include<algorithm>
using namespace std;

const int MAX_N = 1e5 + 1;

struct Node
{
int a;
int b;
int n;
};

bool cmp(Node a, Node b)
{
return a.n < b.n;
};

int par[MAX_N];
int rank1[MAX_N];

void init(int n)
{
for(int i = 0; i < n; ++i)
{
par[i] = i;
rank1[i] = 0;
}
}

int find(int x)
{
if(par[x] == x)
return x;
else
return par[x] = find(par[x]);
}

void unite(int x, int y)
{
x = find(x);
y = find(y);
if(x == y) return ;

if(rank1[x] < rank1[y])
par[x] = y;
else
{
par[y] = x;
if(rank1[x] == rank1[y])
rank1[x]++;
}
}

bool same(int x, int y)
{
return (find(x) == find(y));
}

int main()
{
int n, m;
scanf("%d%d", &n, &m);
Node a[MAX_N];
for(int i = 0; i < m; ++i)
{
scanf("%d%d%d", &a[i].a, &a[i].b, &a[i].n);
}
sort(a, a+m, cmp);
int t = n;
init(n);
for(int i = 0; i < m; ++i)
{
if(!same(a[i].a, a[i].b))
{
unite(a[i].a, a[i].b);
t--;
}
if(t == 1)
{
printf("%d\n", a[i].n);
return 0;
}
}
printf("-1\n");
return 0;
}

洛谷P1197(并查集+逆向思考+邻接表+栈)

中等题,搞了我好久

  • 并查集只加边不减边,所以从所有星球被毁开始计算联通数
  • 由于数据太大,邻接矩阵开不下,只能邻接表(这是我第一次用邻接表)
  • 为了输出简单,我用了栈存储答案
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#include<cstdio>
#include<cstdlib>
#include<stack>
#include<cstring>
using namespace std;

const int MAX_N = 4e5 + 1;

int par[MAX_N]; // 父亲
int rankk[MAX_N]; //树的高度

void init(int n)
{
for(int i = 0; i < n; ++i)
{
par[i] = i;
rankk[i] = 0;
}
}

int find(int x) //查询根
{
if(par[x] == x)
return x;
else
return par[x] = find(par[x]);
}

void unite(int x, int y)
{
x = find(x);
y = find(y);
if(x == y) return ;
if(rankk[x] < rankk[y])
par[x] = y;
else
{
par[y] = x;
if(rankk[x] == rankk[y])
rankk[x]++;
}
}

bool same(int x, int y)
{
return find(x) == find(y);
}

struct Node
{
int data;
Node *next;
};

Node v[MAX_N]; //存起点编号

void initl(int n)
{
for(int i = 0; i < n; ++i)
{
v[i].data = i;
v[i].next = NULL;
}
}

void InsertNode(int a, int b)
{
Node *NewNode = (Node *)malloc(sizeof(Node));
NewNode->data = b;
NewNode->next = v[a].next;
v[a].next = NewNode;
return ;
}

int n, m, ans;
int book[MAX_N];
stack<int> s;

int main()
{
scanf("%d%d", &n, &m);
initl(n);
for(int i = 0; i < m; ++i) //插入边
{
int a, b;
scanf("%d%d", &a, &b);
InsertNode(a, b);
InsertNode(b, a);
}

int k;
scanf("%d", &k);
int deletek[k+1];
memset(deletek, 0, sizeof(deletek));
for(int i = 0; i < k; ++i)
{
scanf("%d", &deletek[i]);
book[deletek[i]] = 1;
}

// for(int i = 0; i < n; ++i)
// {
// Node *p = v[i].next;
// printf("%d: ", i);
// while(p)
// {
// printf("%d ", p->data);
// p = p->next;
// }
// printf("\n");
// }

init(n); //初始化并查集
int ans = n - k;
for(int i = 0; i < n; ++i)
{
Node *p = v[i].next;
if(!book[i]) //没有删去的结点
{
while(p)
{
if(!book[p->data] && !same(i, p->data))
{
unite(i, p->data);
ans--;
}
p = p->next;
}
}
}

s.push(ans);

for(int i = k - 1; i >= 0; --i) //开始加边
{
int a = deletek[i];
Node *p = &v[a];
book[a] = 0;
ans++;
while(p)
{
if(!book[p->data])
{
if(!same(a, p->data))
{
ans--;
unite(a, p->data);
}
}
p = p->next;
}
s.push(ans);
}
while(!s.empty())
{
printf("%d\n", s.top());
s.pop();
}
return 0;
}

洛谷P2024(并查集+思维)

想到其实还好做

  • 不知道每种动物属于哪种,于是开三倍长度的并查集,存储a,a 的天敌,a的天敌的天敌
  • 每次输入a及其天敌及其天敌的天敌都要更新
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#include<cstdio>

const int MAX_N = 1e4 * 5;

int par[MAX_N * 4]; // 父亲
int rank[MAX_N * 4]; //树的高度

void init(int N)
{
for(int i = 0; i < N; ++i)
{
par[i] = i;
rank[i] = 0;
}
}

int find(int x) //查询根
{
if(par[x] == x)
return x;
else
return par[x] = find(par[x]);
}

void unite(int x, int y)
{
x = find(x);
y = find(y);
if(x == y) return ;
if(rank[x] < rank[y])
par[x] = y;
else
{
par[y] = x;
if(rank[x] == rank[y])
rank[x]++;
}
}

bool same(int x, int y)
{
return find(x) == find(y);
}

int ans = 0;
int N, k;

void solve(int T, int X, int Y)
{
if(X > N || Y > N)
{
ans++;
return ;
}
if(T == 1)
{
if(same(X, Y+N) || same(X, Y+2*N))
{
ans++;
return ;
}
else
{
unite(X, Y);
unite(X+N, Y+N);
unite(X+2*N, Y+2*N);
}
}
else
{
if(same(X, Y) || same(X, Y+2*N))
{
ans++;
return ;
}
else
{
unite(X, Y+N);
unite(X+N, Y+N*2);
unite(X+N*2, Y);
}
}
}

int main()
{
scanf("%d%d", &N, &k);
int T, X, Y;
init(3 * N + 3);
for(int i = 0; i < k; ++i)
{
scanf("%d%d%d", &T, &X, &Y);
solve(T, X, Y);
}
printf("%d\n", ans);
return 0;
}

洛谷P1197(带权并查集)

tle了2个点,只有61分,待更

结语

没什么好说的,做得要死要活的,才这么几道简单题,菜是原罪,继续加油吧~