hdu 5326 Work【并查集】

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5326

题意：n个点，给你m有向条边，表示起点是终点的父亲。问子孙节点个数是K的节点有几个？

解法：并查集。

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;int is[110][110];int p[110];int num[110];int n, k;int findd(int x){    if (x == p[x])        return x;    else         return  p[x] = findd(p[x]);}void un(int x, int y){    int a, b;    a = findd(x);    b = findd(y);    if (a == b) return;    p[b] = a;}int main(){    while (scanf("%d%d", &n, &k) != EOF)    {        memset(is, 0, sizeof(is));        memset(num, 0, sizeof(num));        for (int i = 1; i<=n; i++)            p[i] = i;        int x, y;        for (int i = 1; i<n; i++)        {            scanf("%d%d",&x,&y);            un(x,y);            is[x][y] = 1;        }        for (int k = 1; k <= n;k++)        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)            {                if (is[i][k] == 1 && is[k][j] == 1)                {                    is[i][j] = 1;                }            }        for (int i = 1; i <= n; i++)        {            for (int j = 1; j <= n; j++)            {                if (p[j] == p[i] && i != j && is[i][j] == 1)                    num[i]++;            }        }        int ans = 0;        for (int i = 1; i <= n; i++)            if (num[i] == k) ans++;        printf("%d\n", ans);    }}

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