ZOJ-2833*交友問題

一群の人が互いに交友する.M 1 2代表1と2が友人Q 1代表クエリ1の友人数となる.
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
考え方:問題をまとめます.

#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;
	}
}

//            ，       
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("
");
		isfirst = false;
		printf("Case %d:
",++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
",-disjoinset[root1]);
			}

		}
		
	}
}