Find them, Catch them POJ - 1703

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本文介绍了一个经典的并查集算法应用案例，通过处理两个虚拟团伙间成员的关系判断问题，实现了快速查询成员间的归属关系。针对不同类型的指令，如询问成员是否属于同一团伙或明确指出成员不属于同一团伙，文章详细阐述了如何设计并实现算法。

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46665		Accepted: 14367

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

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Link: http://poj.org/problem?id=1703

题意：有两个团伙，给出两种操作：

A X Y：询问 X Y 是否为同伙 ;

D X Y：明确指出X Y为不同伙 ;

思路：典型的并查集思想，但指令给出的是不同伙（根节点不同），将两个团伙分成两个集，但接收到D指令时仍然使用Combine（）合并，使用+n区别开 ;

AC Code：

#include<cstdio>
using namespace std;

const int maxn=500000+50;
int fa[maxn],rank[maxn];

void init(int n){
    for(int i=1;i<=n;i++){
        fa[i]=i;
        rank[i]=0;
    }
}

int find(int x){
    return fa[x]==x?x:fa[x]=find(fa[x]);
}

void myunion(int a,int b){
    a=find(a); b=find(b);
    if(a==b) return ;
    if(rank[a]>rank[b]) fa[b]=a;
    else{
        fa[a]=b;
        if(rank[a]==rank[b]) ++rank[b];
    }

}

bool issame(int a,int b){
    return find(a)==find(b);

}

int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int n,m;
        scanf("%d%d",&n,&m);
        init(n*2);
        while(m--){
            char str[5];
            int a,b;
            scanf("%s%d%d",str,&a,&b);
            //printf("%s %d %d\n",str,a,b);
            if(str[0]=='D'){
                myunion(a,b+n);
                myunion(a+n,b);
            }
            else{
                if(issame(a,b)) printf("In the same gang.\n");
                else if(issame(a+n,b)) printf("In different gangs.\n");
                else printf("Not sure yet.\n");
            }
        }
    }
    return 0;

}