hdu 5326 Work【并查集】

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);
    }
}