ZOJ-2833* 交友问题

一群人相互交友。M 1 2 代表1和2成为朋友 Q 1 代表查询1的朋友数。

Sample Input

3 5
M 1 2
Q 1
Q 3
M 2 3
Q 2
5 10
M 3 2
Q 4
M 1 2
Q 4
M 3 2
Q 1
M 3 1
Q 5
M 4 2
Q 4


Sample Output

Case 1:
2
1
3

Case 2:
1
1
3
1
4


思路：并查集问题。

#include<iostream>
using namespace std;
#include<memory.h>
#include<stdio.h>

int disjoinset[100001];

int find(int * s,int x)
{
	if(s[x] <= 0)
		return x;
	else
		return s[x] = find(s,s[x]); //path compress
}

void unionbysize(int *s,int root1,int root2)
{
	if(s[root1] <= s[root2]) //root1 is bigger
	{
		s[root1] = s[root1] + s[root2];
		s[root2] = root1;
	}
	else
	{
		s[root2] = s[root1] + s[root2];
		s[root1] = root2;
	}
}

//并查集内容 负数代表个数，正数代表父节点
int main()
{
	int N;
	int M;
	char type;
	int mem1;
	int mem2;
	int root1;
	int root2;
	int i = 0;
	bool isfirst = true;

	while(scanf("%d%d",&N,&M) != EOF)
	{
		memset(disjoinset,-1,100001*sizeof(int));	
		if(!isfirst)
			printf("\n");
		isfirst = false;
		printf("Case %d:\n",++i);
		while(M--)
		{
			getchar(); 
			scanf("%c",&type);
			if(type == 'M')
			{
				scanf("%d%d",&mem1,&mem2);
				root1 = find(disjoinset,mem1);
				root2 = find(disjoinset,mem2);
				if(root1 != root2)
					unionbysize(disjoinset,root1,root2);
			}
			else if(type == 'Q')
			{
				scanf("%d",&mem1);
				root1 = find(disjoinset,mem1);
				printf("%d\n",-disjoinset[root1]);
			}

		}
		
	}
}