POJ 1703 Find them, Catch them【带种类的并查集】

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.


题意：就是有两个犯罪团伙，A组给出的x, y而是表明他们不是一伙的.
思路：在原来裸并查集的基础上，多了一个rank[]数组，存放的是father[a]与a的关系，0表示不在一伙，1表示在一伙。father[a]存a的根节点。在状态压缩是也更改rank[]的值为类别转移.
代码如下：

#include<stdio.h>
#include<string.h>
#define N 100005
int father[N], rank[N];
int find(int x)
{
    if(x==father[x])
        return x;
    int t=father[x];
    father[x]=find(t);
    rank[x]=(rank[t]+rank[x])%2;
    return father[x];
} 
int main()
{
    int i, j, k, n, m, T, x, y;
    char ch; 
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        getchar(); 
        for(i=1; i<=n; i++)
            father[i]=i, rank[i]=0;
        for(i=0; i<m; i++)
        {
            scanf("%c%d%d", &ch, &x, &y);
            getchar();
            if(ch=='A')
            {
                if(n==2)    //特殊处理 
                {
                    if(x==y)    printf("In the same gang.\n"); 
                    else    printf("In different gangs.\n");
                    continue; 
                }
                int xx=find(x), yy=find(y);
                if(xx==yy)
                {
                    if(rank[x]==rank[y])    printf("In the same gang.\n");
                    else    printf("In different gangs.\n");
                }
                else
                    printf("Not sure yet.\n");
            }
            else
            {
                int xx=find(x), yy=find(y);
                father[yy]=father[xx];
                rank[yy]=(rank[x]+1+2-rank[y])%2; //类别偏移 
            }
        }
    }
}