poj 1703 Find them, Catch them

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46119		Accepted: 14193

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

/*
* @Author: Lyucheng
* @Date:   2017-07-20 20:33:57
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-07-20 22:12:50
*/
/*
 题意：有两个帮派，两种操作：
    D x,y x和y是属于两个不同的帮派的
    A x,y 询问x和y的关系

 思路：二分染色问题，按照关系建边，如果x,y不在一个联通分量中的话就不能确定关系，如果在一个连通分量重
    那么直接判断与根节点的距离的奇偶
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define MAXN 100005

using namespace std;

int t;
int n,m; 
char str[2];
int x,y;
int bin[MAXN];
int cnt[MAXN];//节点到根结点的距离的奇偶

int findx(int x){
    int cur=x;
    int s=0;
    while(x!=bin[x]){
        if(cnt[x]==1){
            s++;
        }
        x=bin[x];
    }
    bin[cur]=x;
    cnt[cur]=s%2;
    return x;
}

void Union(int x,int y){
    int fx=findx(x);
    int fy=findx(y);
    if(fx!=fy){//不在一个集合内的才合并
        bin[fy]=fx;
        if(cnt[y]==0&&cnt[x]==0)
            cnt[fy]=1;
        else if(cnt[y]==0&&cnt[x]==1)
            cnt[fy]=0;
        else if(cnt[y]==1&&cnt[x]==0)
            cnt[fy]=0;
        else if(cnt[y]==1&&cnt[x]==1)
            cnt[fy]=1;
    }
}

void init(){
    for(int i=0;i<=n;i++){
        bin[i]=i;
    }
    memset(cnt,0,sizeof cnt);
}

int main(){ 
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=0;i<m;i++){
            scanf("%s%d%d",str,&x,&y);
            if(str[0]=='A'){//询问
                int fx=findx(x);
                int fy=findx(y);
                if(fx!=fy){
                    puts("Not sure yet.");
                }else{
                    if(cnt[x]==cnt[y]){
                        puts("In the same gang.");
                    }else{
                        puts("In different gangs.");
                    }
                }
            }else{//关系
                Union(x,y);
            }
        }
    }
    return 0;
}