POJ 1703 Find them, Catch them

題目：

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 31232		Accepted: 9621

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.

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解題思路：

分組並查集(種類並查集).建立兩個集合A和B，A中存放和元素a在同一個幫派的元素，B中存放和元素a在不同幫派（一共兩個幫派）的元素。對於'D'操作，我們只需把a和b+n放到一個集合，a+n和b放到一個集合即可，因爲a和b不在一個幫派，所以a和b+n在一個幫派，同理，a+n和b在一個幫派。對於‘A’操作，我們先判斷a和b是否在同一個幫派，即看a和b或者a+n和b+n是否在同一個集合，再看兩個元素是否在不同幫派，即看a+n和b或者a和b+n是否在同一個集合，剩下的情況是不確定。

代碼：

#include <cstdio>
#include <cstring>

const int MAXN = 1e5 + 10;
int set[MAXN<<1];

int find(int p)
{
    if(set[p] < 0) return p;
    return set[p] = find(set[p]);
}

void join(int p, int q)
{
    p = find(p), q = find(q);
    if(p != q) set[p] = q;
}

int main()
{
    int t, n, m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        memset(set, -1, sizeof(set));
        while(m--)
        {
            char s[2];
            int a, b;
            scanf("%s%d%d", s, &a, &b);
            switch(s[0])
            {
                case 'D':
                {
                    join(a, b+n);
                    join(b, a+n);
                    break;
                }
                case 'A':
                {
                    if(find(a)==find(b) || find(a+n)==find(b+n))
                    {
                        printf("In the same gang.\n");
                    }
                    else if(find(a)==find(b+n) || find(a+n)==find(b))
                    {
                        printf("In different gangs.\n");
                    }
                    else
                    {
                        printf("Not sure yet.\n");
                    }
                    break;
                }
            }
        }
    }
}