POJ 1703 并查集

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 41878		Accepted: 12868

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.

题意：城市里有两伙罪犯，如果是D[ a ][ b ]，则代表a和b是不同伙的罪犯；如果是D[ a ][ b ]，则输出a和b的关系。

思路：并查集，是那道超经典并查集例题 “ 食物链 ” 的简化版，当时做那道 “ 食物链 ” 时了解了两种思路：一是来自watashi菊苣译得那本《挑战程序设计竞赛》的用三个集合来代表三种组织；二是网上流传的方法，用一个带权并查集来表示，其中权值代表各结点和祖先的关系。

这道题我用的是watashi译书中的方法，但由于有两个团伙，所以开数组时要X2，1——N表示一个组织，N+1——2N表示另一个组织。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
int fa[200005],rank[200005];
int find(int x)
{
    if(x==fa[x])
        return x;
    else
        return fa[x]=find(fa[x]);
}
void unite(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)
        return;
    if(rank[x]<rank[y])
        fa[x]=y;
    else
    {
        fa[y]=x;
        if(rank[x]==rank[y])
            rank[x]++;
    }
}
int main()
{
    //freopen("D:\\in.txt","r",stdin);
    int T;
    char ch[2];
    scanf("%d",&T);
    while(T--)
    {
        int m,n;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=2*n;i++)
        {
            fa[i]=i;
            rank[i]=0;
        }
        int x,y,cc;
        while(m--)
        {
            scanf("%s%d%d",ch,&x,&y);
            getchar();
            if(ch[0]=='D')
            {
                unite(x,y+n);
                unite(x+n,y);
            }
            else if(ch[0]=='A')
            {
                if(find(x)!=find(y)&&find(x)!=find(y+n))
                    printf("Not sure yet.\n");
                else if(find(x)==find(y)||find(x+n)==find(y+n))
                    printf("In the same gang.\n");
                else if(find(x)==find(y+n)||find(x+n)==find(y))
                    printf("In different gangs.\n");
            }
        }
    }
    return 0;
}