POJ 1703 Find them, Catch them （并查集）

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23816		Accepted: 7130

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

 

 

 

 

并查集的应用，用于判断种类，对2取模就可以了。

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

const int MAXN=100010;
int F[MAXN];
int val[MAXN];
int find(int x)
{
    if(F[x]==-1)return x;
    int tmp=find(F[x]);
    val[x]=(val[x]+val[F[x]])%2;
    return F[x]=tmp;
}
void bing(int x,int y)
{
    int t1=find(x);
    int t2=find(y);
    if(t1!=t2)
    {
        F[t1]=t2;
        val[t1]=(val[y]-val[x]+1)%2;
    }
}
int main()
{
    int T;
    char str[10];
    int u,v;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(F,-1,sizeof(F));
        memset(val,0,sizeof(val));
        while(m--)
        {
            scanf("%s%d%d",&str,&u,&v);
            if(str[0]=='A')
            {
                if(find(u)!=find(v))
                {//题目说两个集团至少有一个人，所以N==2的时候单独考虑，但是不考虑这个也可以AC，估计没有这样的数据
                    if(n==2)printf("In different ganges.\n");
                    else printf("Not sure yet.\n");
                }
                else
                {
                    if(val[u]!=val[v])printf("In different gangs.\n");
                    else printf("In the same gang.\n");
                }
            }
            else
            {
                bing(u,v);
            }
        }
    }
    return 0;
}