poj 1703 Find them, Catch them

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 33780		Accepted: 10432

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

 

1.并查集

2.

逻辑判断1：a与fa的关系与fa与f[a]的关系相同，也就是对于fa，a和f[a]等价，所以 a和f[a]一定在同一组织；a与fa的关系与fa与f[a]的关系不同（相反），也就是对于fa，a和f[a]不等价，所以a和f[a]一定不在同一组织。

逻辑判断2：a、b必然不在同一组织中，4种情况

 

//pojFind them, Catch them  1703
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<queue>
using namespace std;
int f[100005];
bool c[100005];
void Init(int n){
    int i;
    for(i=0;i<=n;i++){
        f[i]=i;
        c[i]=true;//true表示与父节点在同一组织内
    }
}
int find_set(int a){
    int fa=f[a];
    if(f[a]!=a){
        f[a]=find_set(f[a]);
    }
    c[a]=(c[fa]==c[a])?true:false;//逻辑判断1
    return f[a];
}
void Union(int a,int b,int fa,int fb){
    f[fa]=fb;
    c[fa]=(c[a]==c[b])?false:true;//逻辑判断2
}
int main(){
    //freopen("D:\\INPUT.txt","r", stdin);
    int T,n,m,i,a,b;
    char ch;
    cin>>T;
    while(T--){
        scanf("%d %d\n",&n,&m);//cin>>n>>m;
        Init(n);
        for(i=0;i<m;i++){
            // cin 会导致TLE
            scanf("%c %d %d\n",&ch,&a,&b);
            //cin>>ch;
            //cout<<ch<<endl;
            //cin>>a>>b;//scanf("%c %d %d\n",&ch,&a,&b);//cin>>ch>>a>>b;
            int fa,fb;
            fa=find_set(a);
            fb=find_set(b);
            if(ch=='A'){
                if(fa==fb){
                    if(c[a]==c[b]){
                        cout<<"In the same gang."<<endl;
                    }
                    else{
                        cout<<"In different gangs."<<endl;
                    }
                }
                else{
                    cout<<"Not sure yet."<<endl;
                }
            }
            else{
                Union(a,b,fa,fb);//if(fa!=fb)
            }
        }
    }
    return 0;
}