PKU 1703 【并查集】

题目：

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12040		Accepted: 3493

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 

这题目和黑书81页《团伙》那题目很类似，用e[i]记录和集合i敌对的集合，主要思想是和a对立的元素一定是属于同一个集合，所以每次输入

D I J时，就把e[i]和j合并，e[j]和i合并，在查询时，如果a和b属于同一个集合则same,如果a等于集合e[b]，则一定是different，否则就是

not sure.

 #include <iostream>

#include <cstring>
#include <cstdio>
using namespace std;
const int N = 100000+40;
int fa[N];
int e[N];
int rank[N];
int n,m;
void init()
{
    for(int i=0;i<=n+10;i++)
    {
        fa[i] = i;
        e[i] = i;
        rank[i] = 0;
    }
}
int getF(int x)
{
    if(x==fa[x])return fa[x];
    return fa[x] = getF(fa[x]);
}
void merge(int x,int y)
{
    int fx = getF(x);
    int fy = getF(y);
    if(fx!=fy)
    {
        if(rank[fx]<rank[fy])
        {
            fa[fx] = fy;
            rank[fy]++;
        }
        else
        {
            fa[fy] = fx;
            rank[fx]++;
        }
    }
}
char str[10];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        init();
        for(int i=0;i<m;i++)
        {
            int ta,tb;
            int tx,ty;
            scanf("%s%d%d",str,&ta,&tb);
            ta+=2;tb+=2;
            tx = getF(ta);ty = getF(tb);
            if(str[0]=='A')
            {
                if(tx==ty)
                {
                    printf("In the same gang.\n");
                }
                else if(tx==getF(e[ty]))
                {
                    printf("In different gangs.\n");
                }
                else
                {
                   printf("Not sure yet.\n");
                }
            }
            else
            {
                if(e[ty]!=ty)
                {
                    merge(tx,e[ty]);
                }
                else
                {
                    e[ty] = tx;
                }
                if(e[tx]!=tx)
                {
                    merge(ty,e[tx]);
                }
                else
                {
                    e[tx] = ty;
                }
            }
        }
    }
    return 0;
}