题目不难理解，A判断2人是否属于同一帮派，D确认两人属于不同帮派。于是需要一个数组r[]来判断父亲节点和子节点的关系。具体思路可参考

#include 
     
      #include 
      
       #include 
       
        using namespace std;const int maxn=100005;int f[maxn];int r[maxn];int Find_f(int x){    if (x != f[x])    {        int t=f[x];        f[x] = Find_f(f[x]);        r[x]=(r[x]+r[t])%2;        return f[x];    }    return f[x];}void Union(int x,int y){    int fx=Find_f(x);    int fy=Find_f(y);    f[fx]=fy;    r[fx]=(r[x]+1+r[y])%2;}int main(){    //freopen("in.txt","r",stdin);    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)        {            f[i]=i;            r[i]=0;        }        char c;        int crime1,crime2;        while(m--)        {            getchar();            scanf("%c%d%d",&c,&crime1,&crime2);            //printf("%c\n",c);            if(c=='D')                Union(crime1,crime2);            else if(c=='A')            {                if(Find_f(crime1)==Find_f(crime2))                    if(r[crime1]!=r[crime2])                        printf("In different gangs.\n");                    else                        printf("In the same gang.\n");                else                    printf("Not sure yet.\n");            }        }    }    return 0;}