poj1703

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38075		Accepted: 11714

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
题意：
有两个帮派，A x y 表示询问x 与 y 是否是一个帮派，此时输出回答是或不是 或者不确定 ，D x y 表示x y 不是一个帮派，
#include <iostream>
#include <stdio.h>
using namespace std;
const int maxn=1e5+10;
int par[maxn];
int ans[maxn];
void init(int n)
{
    for(int i=0; i<=n; i++)
    {
        par[i]=i;
        ans[i]=0;
    }
}
int find(int i)
{
    if(i==par[i])
        return i;
    else
    {
        int tr=find(par[i]);
        ans[i]=ans[i]^ans[par[i]];
        return par[i]=tr;
    }

}
void unite(int x,int y)
{
    int  nx=find(x);
    int ny=find(y);
    par[nx]=ny;
    ans[nx]=~(ans[x]^ans[y]);///~  按位取反 类别偏移
 
}
bool same(int x,int y)
{
    return find(x)==find(y);
}
int main()
{
    int T,n,m;
    char str;
    int x,y;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init(n);
        for(int i=0; i<m; i++)
        {
            cin>>str;
            scanf("%d%d",&x,&y);
            getchar();
            if(str=='A')
            {
                if(n==2)
                {
                    printf("In different gangs.\n");
                    continue;
                }
                else if(same(x,y))
                {
                    if(ans[x]==ans[y])
                        printf("In the same gang.\n");
                    else
                        printf("In different gangs.\n");
                }
                else
                {
                    printf("Not sure yet.\n");
                }
            }
            if(str=='D')
            {
                unite(x,y);
            }
        }
    }
    return 0;
}