poj1703 Find them Catch them

http://www.elijahqi.win/2018/01/12/poj1703-find-them-catch-them/ ‎

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
题意：每次给定两个人 认为他们是敌人 并且题目保证只有两个团伙 然后多组询问 询问他们是朋友还是敌人还是暂时关系不明确
设立e数组 代表每个数的敌人分别是谁 但这个敌人我不需要全部记录 我只需要 选一个敌人作为代表记录即可 每次说这两个人是敌人 那么就分别把他们和对方的敌人连起来 表示他们是朋友 同时看情况补充到对方的敌人数组中 询问的时候 如果是朋友直接并查集判断即可 如果是敌人 判断x的敌人和y的关系即可 如果都不属于上面给定的情况 说明他们暂时还没有关系

#include
#define N 110000
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while(ch<'0'||ch>'9') ch=gc();
    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x;
}
int T,fa[N],n,m,e[N];
inline int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
inline void merge(int x,int y){
    int fx=find(x),fy=find(y);
    if (fx!=fy) fa[fx]=fy;
}
int main(){
    freopen("poj1703.in","r",stdin);
    T=read();
    while(T--){
        n=read();m=read();
        for (int i=1;i<=n;++i) fa[i]=i,e[i]=0;
        for (int i=1;i<=m;++i){
            char ch=gc();while(ch!='A'&&ch!='D') ch=gc();
            if (ch=='A'){
                int x=read(),y=read();
                if (find(x)==find(y)) {puts("In the same gang.");continue;}
                if (find(x)==find(e[y])) {puts("In different gangs.");continue;}
                puts("Not sure yet.");
            }else{
                int x=read(),y=read();
                if (!e[x]) e[x]=y;else merge(e[x],y);
                if (!e[y]) e[y]=x;else merge(e[y],x);
            }
        }
    }
    return 0;
}