这种类型的题目以前见过，今天第一次写，具体过程，还要慢慢理解。
1 #include <cstring> 2 #include <cstdio> 3 #include <string> 4 #include <iostream> 5 #include <algorithm> 6 #include <vector> 7 using namespace std; 8 int o[100001]; 9 int flag[100001]; 10 int find(int x) 11 { 12 if (x == o[x]) return x; 13 int t = find(o[x]); 14 flag[x] = (flag[o[x]] + flag[x]) % 2; 15 return o[x] = t; 16 } 17 void merge(int x,int y) 18 { 19 int tx,ty; 20 tx = find(x); 21 ty = find(y); 22 if(tx != ty) 23 { 24 o[tx] = ty; 25 flag[tx] = (flag[x]+flag[y]+1)%2; 26 } 27 } 28 int judge(int x,int y) 29 { 30 int tx,ty; 31 tx = find(x); 32 ty = find(y); 33 if(tx != ty) 34 return -1; 35 else 36 return (flag[x] == flag[y]); 37 } 38 int main() 39 { 40 int T,sv,ev,i,n,m; 41 char ch; 42 scanf("%d",&T); 43 while(T--) 44 { 45 scanf("%d%d%*c",&n,&m); 46 for(i = 1;i <= n;i ++) 47 { 48 flag[i] = 0; 49 o[i] = i; 50 } 51 for(i = 0;i < m;i ++) 52 { 53 scanf("%c%*c%d%d%*c",&ch,&sv,&ev); 54 if(ch == 'A') 55 { 56 int temp = judge(sv,ev); 57 if(temp == -1) 58 printf("Not sure yet. "); 59 else if(temp == 0) 60 printf("In different gangs. "); 61 else 62 printf("In the same gang. "); 63 } 64 else 65 { 66 merge(sv,ev); 67 } 68 } 69 } 70 return 0; 71 }