POJ 2524 ：Ubiquitous Religions

标签：poj

Description

Input

Output

Sample Input

10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Sample Output

Case 1: 1
Case 2: 7

额。。难道题单错了？。。。我一看就是并查集。。所以就这么水过了。。

并查集一A水过

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define M 100500
int p[M];
int n, m;

void start()
{
    for(int i=1; i<=n; i++)
        p[i] = i;
}

int find(int x)
{
    return p[x] == x ? x : p[x] = find( p[x] );
}

void Kruskal(int x, int y)
{
    int xx = find(x);
    int yy = find(y);
    if(xx!=yy)
        p[yy] = xx;
}

int main()
{
    int cas = 0;
    while(scanf("%d%d", &n, &m) &&n &&m)
    {
        cas++;
        start();
        int ans = 0;
        for(int i=1; i<=m; i++)
        {
            int x; int y;
            scanf("%d%d", &x, &y);
            Kruskal(x, y);
        }
        for(int i=1; i<=n; i++)
        {
            if( p[i]==i )
                ans++;
        }
        printf("Case %d: %d\n", cas, ans);
    }

    return 0;
}

POJ 2524 ：Ubiquitous Religions,布布扣,bubuko.com

POJ 2524 ：Ubiquitous Religions

标签：poj

原文地址：http://blog.csdn.net/u013487051/article/details/37830479