题意：

在这个城市里有两个黑帮团伙，现在给出N个人，问任意两个人他们是否在同一个团伙
1.输入D x y代表x于y不在一个团伙里
2.输入A x y要输出x与y是否在同一团伙或者不确定他们在同一个团伙里

题目：

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.

Sponsor

分析：

这道题用的是种类并查集：并查集把给出的人分成几个集合，每个集合之间的人的关系不确定，对同一个集合保存和本人不为同一队的人，本着敌人的敌人便是朋友的原则，用并查集同一集合为同一队，不同集合为不同队。说的我自己都绕晕了，23333
1.首先特殊解很重要：当N=2时他们属于不同的帮派，因为题目有说两个帮派至少有一个人。
2.
（1）只要输入D，就将a,b两个合并，归在同一集合，并将改他们的关系。
（2）输入A的时候判断a,b是否合并过，如果两个不属于同一个集合的话就不能确定他们是否在同一个帮派。
（3）若合并过，即前面已经出现过，则有确定关系即是否在同一帮派。此时只要判定其dp【】值是否相同，即为一个队，为同一帮派，否则不在。
原因：我在把一个集合合并到另一个集合时，把x根节点的dp变成和y根节点dp相对的，（每次连儿子保证和父亲不是一个帮派，同时更新父亲）然后在查找的时候要修改dp值（注意回溯），因为生成的树每一层和隔层的dp值是相对的（0和1），因为dp的值只能为0和1（只有两个帮派），所以类别偏移用位运算
（4）若我太啰嗦，可参照 如下AC代码，有步骤详解：

#include <stdio.h>
#define M 100010
int dp[M],f[M];
int i,n,m,a,b;
int t;
int find(int x)
{int num;if(f[x]!=x){num=find(f[x]);dp[x]=dp[x]^dp[f[x]];//类别偏移可以用按位异或运算，当两对应的二进位相异时，结果为1。dp值是相对的（0和1）return f[x]=num;}return x;
}
void dfs(int x,int y)
{int u=find(x);int v=find(y);f[u]=v;///dp[u]=~(dp[y]^dp[x]);//类别偏移，用按位取反运算，即dp值是相对的（0和1）dp[u]=1^dp[y]^dp[x]; /**更新x的父节点跟y（y的父节点）的关系,效果与按位取反相同.每次连儿子保证和父亲不是一个帮派，同时更新父亲，然后find时候注意回溯一下。*/
}
int main()
{scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0; i<=n; i++){dp[i]=0;//0表示同派f[i]=i;}for(i=1; i<=m; i++){char s[5];scanf("%s%d%d",s,&a,&b);if(s[0]=='D')//只要输入D，就将a,b两个合并，并将改他们的关系dfs(a,b);else//,输入A的时候判断a,b是否合并过，没有输出无法确定，合并过再判断是不是同一派。{if(n==2)    //特殊解printf("In different gangs.\n");else if(find(a)==find(b))///前面已经出现过，则有确定关系即是否在同一帮派{if(dp[a]==dp[b])printf("In the same gang.\n");elseprintf("In different gangs.\n");}elseprintf("Not sure yet.\n");}}}return 0;
}