POJ 1703 Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 34672		Accepted: 10691

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.

题解:并查集经典题，很不错。本来觉得是把并查集的过程反过来，并用一个bool数组记录i这个人有没有操作过，操作过才能知道他的阵营，否则不能，但是后来发现这显然是错了，还有很多不能确定。后来看了看别人的做法，突然发现当D时，只需要把x和y的敌人合并，y和x的敌人合并，而i的敌人只需要用一个数组记录一下就可以了。而查询时，如果find(x)与y在一个联通块则是一个，如果和对方的敌人是一个联通块怎不是一个，否则则无法判断。
CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 100000 + 10

using namespace std;

int n, q, F[MAX_N], E[MAX_N];

int find(int x){
    if(F[x] == x) return x;
    return F[x] = find(F[x]);
}

void Union(int x,int y){
    x = find(x), y = find(y);
    if(x == y) return;
    F[y] = x;
}

int main(){
    int T; scanf("%d", &T);
    while(T --){
        scanf("%d%d", &n, &q);
        REP(i, 1, n) F[i] = i, E[i] = 0;
        
        char s[3]; int x, y;
        while(q --){
            scanf("%s", s);
            if(s[0] == 'A'){
                scanf("%d%d", &x, &y);
                if(find(x) == find(y)) printf("In the same gang.\n");
                else if(find(x) == find(E[y])) printf("In different gangs.\n");
                else printf("Not sure yet.\n");
            }
            else if(s[0] == 'D'){
                scanf("%d%d", &x, &y);
                if(!E[x] && !E[y]) E[x] = y, E[y] = x;
                else if(!E[x]) E[x] = y, Union(x, E[y]);
                else if(!E[y]) E[y] = x, Union(y, E[x]);
                else if(E[x] && E[y]) Union(x, E[y]), Union(y, E[x]);
            }
        }
    }
    return 0;
}