poj--1703--Find them, Catch them（并查集巧用）

Find them, Catch them

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

SubmitStatus

Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:

1. D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

Sample Input

1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4

Sample Output

Not sure yet.
In different gangs.
In the same gang.

Source

POJ Monthly--2004.07.18

输出少了一个“.”wa了四次-.-||
题意：有两个帮派，现在给了n个人m个关系，D：关系表示两个人不在一个帮派，A：查询两个人是否在一个帮派，不确定的话输出“Not sure yet.“，同一个输出”In the same gang.“，不在一个输出”In different gangs.“。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int pre[100010],rank[100010];
int find(int x)
{
	if(pre[x]==x)
	return x;
	else
	{
		int temp=pre[x];
		pre[x]=find(pre[x]);
		rank[x]=(rank[x]+rank[temp])%2;//将x变为和temp一样的帮派 
		return pre[x];
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(rank,0,sizeof(rank));//rank表示层数，两层表示不一样的帮派 
		char op[2];
		int x,y;
		int m,n;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		pre[i]=i;
		for(int i=0;i<m;i++)
		{
			scanf("%s%d%d",op,&x,&y);
			int fx=find(x);
			int fy=find(y);
			if(op[0]=='D')
			{
				pre[fy]=fx;
				rank[fy]=(rank[x]-rank[y]+1)%2;
				continue;
			}
			else
			{
				if(fx!=fy)//根节点不一样应该就是尚未分配，所以不确定 
				printf("Not sure yet.
");
				else if(rank[x]!=rank[y])//层数不一样就不在一个帮派 
				printf("In different gangs.
");
				else if(rank[x]==rank[y])//层数一样并且根节点相同 
				printf("In the same gang.
");
			}	
		}
	}
	return 0;
}