POJ - 2524 Ubiquitous Religions（简单并查集）

标签：acm   并查集   

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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

Case 1: 1
Case 2: 7

简单并查集，求无向图连通分量。

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int MAXN = 50000 + 50;
int parent[MAXN];

int n, m;

void make_set()
{
	for (int i = 0; i <= n; i++)
		parent[i] = -1;
}

int find_set(int t)
{
	if (parent[t] == -1)
		return t;
	else
		return parent[t] = find_set(parent[t]);
}

void union_set(int a, int b)
{
	int t1 = find_set(a);
	int t2 = find_set(b);
	if (t1 != t2)
		parent[t2] = t1;
}

int main()
{
	int casen;
	casen = 1;
	while (scanf("%d%d", &n, &m) != EOF&&n)
	{
		int a, b;
		make_set();
		for (int i = 0; i < m; i++)
		{
			scanf("%d%d", &a, &b);
			union_set(a, b);
		}
		int sum = 0;
		for (int i = 1; i <= n; i++)
		{
			if (parent[i] == -1)
				sum++;
		}
		printf("Case %d: ", casen);
		casen++;
		printf("%d\n", sum);
	}
}

POJ - 2524 Ubiquitous Religions（简单并查集）

标签：acm   并查集   

原文地址：http://blog.csdn.net/qq_18738333/article/details/47978147