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：
给出A和D两种操作，A询问a，b两个数是否在一个集团（注意不是集合，若a,b不在一起，c，d不在一起，但不能说明ac,ad,bc,bd,是否在哦一起）
D 操作a，b不在一个集团；
思路：
用种类并查集，用ran[]数组表示子结点与父结点的关系，1 表示为与父结点在一个集团，0表示不再一个集团；
由于自己本身一定在一个集团所以将ran数组初始化为1；
#include<stdio.h>
#include<string.h>
#define MAX 100005
using namespace std;
int n,m;
int pre[MAX];
int ran[MAX];
int init(){//初始化
int i;
for(i=0;i<=n;i++){
pre[i]=i;
ran[i]=1;
}
}
int Find(int x){
if(pre[x]==x) return x;
int fx=pre[x];
pre[x]=Find(pre[x]);//路径压缩
ran[x]=(ran[x]+ran[fx]+1)%2;//关于这个关系式大家画个表格找一下关系
return pre[x];
}
void mix(int a,int b){
int fa=Find(a),fb=Find(b);
if(fa!=fb){
pre[fb]=fa;
ran[fb]=(ran[a]+ran[b])%2;//关于这个关系式大家画个表格找一下关系
}
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
init();
int i;
char s[5];
for(i=0;i<m;i++){
int a,b;
scanf("%s%d%d",s,&a,&b);
if(s[0]=='A'){
int fa=Find(a),fb=Find(b);
if(fa==fb){
if(ran[a]==ran[b]) printf("In the same gang.\n");
else printf("In different gangs.\n");
}
else printf("Not sure yet.\n");
}
else{
mix(a,b);
}
}
}
return 0;
}
