Find them, Catch them POJ - 1703

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.

题目意思：两个帮派，d操作告诉你x,y属于不同阵营。
题目思路：和上题有相似之处，用x+n表示x的敌对阵营，y+n表示y的敌对阵营（给的每句话都是真话）。 操作A，读入两个X,Y。如果x,y同根，说明属于同一阵营。如果x,y+n同根或者x+n,y同根，说明xy属于相反阵营，剩下就是不能判断的情况。

#include <stdio.h>
int parent[200000+5];
int find(int x)
{
    return (parent[x]==x)?x:parent[x]=find(parent[x]);
}
int same(int x, int y)
{
    return find(x)==find(y);
}
void unite(int x,int y)
{
    x=find(x);
    y=find(y);
    if(x==y)  return ;
    else parent[y]=x;
}
void init(int n)
{
    for(int i=1;i<=n;i++)
        parent[i]=i;
}
int main(void)
{
    int n,m;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char str[2];
        scanf("%d%d",&n,&m);
        init(2*n);
        for(int i=1;i<=m;i++)
        {
            int num1,num2;
            scanf("%s",str);
            scanf("%d%d",&num1,&num2);
            if(str[0]=='D')
            {
                unite(num1,num2+n);
                unite(num1+n,num2);
            }
            else
            {
                if(same(num1,num2)==1)
                    printf("In the same gang.\n");
                else if(same(num1,num2+n)==1 || same(num1+n,num2)==1)
                    printf("In different gangs.\n");
                else
                    printf("Not sure yet.\n");
            }
        }
    }
}