POJ 1703 Find them, Catch them

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21108		Accepted: 6259

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

Sample Output

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

Source

POJ Monthly--2004.07.18

//有了食物链那题的基础、这题就思路清晰了

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define Y 100003
using namespace std;
int f[Y],r[Y];
bool p[Y];
int Find(int x)
{
    if(x!=f[x])
    {
        int tf=f[x];
        f[x]=Find(f[x]);
        p[x]=!(p[x]^p[tf]);//压缩路径的规律
        return f[x];
    }
    return x;
}
void union_set(int x,int y)
{
    int rx=Find(x),ry=Find(y);
    if(r[rx]>r[ry])
     {
         f[ry]=rx;
         p[ry]=(p[x]+p[y])%2;//合并的规律
     }
     else if(r[rx]<r[ry])
          {
              f[rx]=ry;
              p[rx]=(p[x]+p[y])%2;
          }
          else
          {   f[ry]=rx;r[rx]++;
              p[ry]=(p[x]+p[y])%2;
          }
}
int main()
{
    int T;
    int N,M;
    char c;
    int x,y,r1,r2;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++)
           f[i]=i,r[i]=p[i]=1;
        while(M--)
        {
            getchar();
            scanf("%c",&c);
            scanf("%d%d",&x,&y);
            if(N==2&&c=='A'&&x!=y)//特殊情况 2个人 A 1 2是可以确定的
            {
                printf("In different gangs.\n");continue;
            }
            if(c=='D')
              union_set(x,y);
            else
             {
                  r1=Find(x);
                  r2=Find(y);
                 if(r1!=r2)
                   printf("Not sure yet.\n");
                 else
                 {
                     if(p[x]==p[y])
                       printf("In the same gang.\n");
                    else
                       printf("In different gangs.\n");//到这我不得不说句：你妹的！！！，少写了个s，WA了几次，晕
                 }
             }
        }
    }
    return 0;
}