【POJ】Find them, Catch them

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 62168		Accepted: 18835

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

【思路】

  使用并查集。但是并查集内相同父节点认为是一个集合内，这道题的题意是给我们一对数，这对数不在一个集合内。

  例子：

  A B

  B C

  A 和 B 不在一个集合内， B 和 C 不在一个集合内， 可以推导出A和C在一个集合内。如果我们直接使用并查集的话。A，B，C

  回使A，B，C 在相同的集合内。这样就找不到A，B，C的关系了。

  解决办法是：

  A 和 B + n 在一个集合内

  B 和 A + n 在一个集合内

  C 和 B + n 在一个集合内

  B 和 C + n 在一个集合内

  这样 查询 A 和 B 是否在一个集合的时候，我们使用 find（A， B+n) 和 find(A + n, B) 

  在判断 A 和 C 是否一个集合的时候， 我们通过 A 和 B + n 和 C 和 B + n 推导出 A 和 C 在一个团伙里。

【代码】

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e5 + 10;
int s[N];
int t, n, m;
char c;
void init() {
    memset(s, 0, sizeof(s));
    for(int i = 1; i <= n * 2; i++) {
        s[i] = i;
    }
}
int find(int x) {
    if(s[x] != x) s[x] = find(s[x]);

    return s[x];
}
void union_set(int x, int y) {
    x = find(x);
    y = find(y);
    if(x != y ) s[x] = s[y];
}

bool same(int x, int y ) {
    return find(x) == find(y);
}
int main() {
    cin >> t;
    while(t--) {
        scanf("%d%d", &n, &m);
        init();

        getchar();
        for(int i = 0; i < m; i++) {
            int x, y;
            scanf("%c %d %d", &c, &x, &y);
            getchar();
            if(c == 'A') {
                if(n == 2) {
                    printf("In different gangs.\n");
                } else if( same(x, y) ) {
                    printf("In the same gang.\n");
                } else if( same(x, y + n) || same(y, x + n)) {
                    printf("In different gangs.\n");
                } else {
                    puts("Not sure yet.");
                }
            } else {
                union_set(x, y + n);
                union_set(y, x + n);
            }
        }
    }
    return 0;
}