POJ 1703 - Find them, Catch them

POJ 1703 - Find them, Catch them
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.

题意：一个城市有两个帮派，现在公安局掌握了一些资料，但是只知道A,B两个属于不同的帮派。要你根据所给的信息判别任意两个人的关系，是属于同一个帮派还是属于不同的帮派，还是他们的关系不能确定了？

解题思路：
食物链的简化版
A x y 不能确定xy是否是同一帮派
D x y 确定xy属于不同帮派
x表示帮派A x+n表示帮派B

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
const int N = 100000;
const int MAX = 2*N+16;

int fa[MAX], deep[MAX];

void init()
{
    memset(fa, -1, sizeof(fa));
    memset(deep, 0, sizeof(deep));
}

int find(int x)
{
    if (fa[x] == -1) return x;
    return fa[x] = find(fa[x]);
}

void unite(int x, int y)
{
    x = find(x), y = find(y);
    if (x == y) return;
    if (deep[x]<deep[y])
        fa[x] = y;
    else
    {
        fa[y] = x;
        if (deep[x] == deep[y])
            deep[x]++;
    }
}

bool same(int x, int y)
{
    return find(x) == find(y);
}


int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
    int n, m;
    int x, y;
    char c[2];//这里如果定义char c 会出现Runtime Error
        if (T >= 20) break;
        scanf("%d%d", &n, &m);
        init();
        //cin>>n>>m;
        if (n>N || m>N) break;
        while (m--) {
            scanf("%s%d%d", c, &x, &y);
            //cin >> c >> x >> y;
            if (c[0] == 'D')
            {
                unite(x, y + n);
                unite(x + n, y);
            }
            else {
                if (same(x, y) || same(x + n, y + n))
                {
                    printf("In the same gang.\n");
                    continue;
                }
                if (same(x, y + n) || same(x + n, y))
                {
                    printf("In different gangs.\n");
                    continue;
                }
                printf("Not sure yet.\n");
            }
        }

    }
    return 0;
}