POJ 1703 Find them, Catch them（确定元素归属集合的并查集）

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 52925		Accepted: 16209

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

题目意思：

现在又两个犯罪团伙

罪犯编号1到n，现在又两种操作：

1.D x y 告诉你编号x的罪犯和编号y的罪犯属于不同犯罪团伙

2.A x y 问你x和y是不是属于同一个犯罪犯罪团伙，三种情况，是同一个，不是同一个，不确定

 

分析：

union_set(x, y)同属于第一个团伙

union_set(x+n,y+n)同属于第二个团伙

union_set(x+n, y)表示x属于第二个团伙，y属于第一个团伙

union_set(x, y+n)表示x属于第一个团伙，y属于第二个团伙；

每次告诉你x和y属于不同团伙，有两种情况，x属于1，y属于2

x属于2，y属于1，所以需要这样进行两次合并

确定是否属于相同团伙的话

x与y的根结点相同或者x+n与y+n的根结点相同 都属于相同犯罪团伙

x+n与y根结点相同或者x与y+n的根结点相同，都属于不同犯罪团伙

其他情况就不能确定了

这个思路很巧妙，也比较简单，不用具体确定x属于1还是2

code：

#include<queue>
#include<set>
#include<cstdio>
#include <iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define max_v 100005
int pa[max_v*];
int rk[max_v*];
int n,m;
void init()
{
    for(int i=; i<=*n; i++)
    {
        pa[i]=i;
        rk[i]=;
    }
}
int find_set(int x)
{
    if(x!=pa[x])
        pa[x]=find_set(pa[x]);
    return pa[x];
}
void union_set(int x,int y)
{
    x=find_set(x);
    y=find_set(y);
    if(x==y)
        return ;
    if(rk[x]>rk[y])
        pa[y]=x;
    else
    {
        pa[x]=y;
        if(rk[x]==rk[y])
            rk[y]++;
    }
}
int f(int x,int y)
{
    return find_set(x)==find_set(y);
}
int main()
{
    int t;
    scanf("%d",&t);
    int x,y;
    char str[];
    while(t--)
    {
        scanf("%d %d",&n,&m);
        init();

        for(int i=; i<m; i++)
        {
            scanf("%s %d %d",str,&x,&y);
            if(str[]=='D')
            {
                union_set(x+n,y);
                union_set(x,y+n);
            }
            else if(str[]=='A')
            {
                if(f(x,y)||f(x+n,y+n))
                    printf("In the same gang.\n");
                else if(f(x+n,y)||f(x,y+n))
                    printf("In different gangs.\n");
                else
                    printf("Not sure yet.\n");
            }
        }
    }
    return ;
}