poj1703 Find them, Catch them

并查集。

这题错了不少次才过的。

分析见代码。

http://poj.org/problem?id=1703

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int maxn = 1e5 + ;
 const char str1[] = "Not sure yet.";
 const char str2[] = "In different gangs.";
 const char str3[] = "In the same gang.";
 int fa[maxn], belong[maxn], nex[maxn];
 int n, m, u, v, k;
 char op;

 int father(int u){
     //WHAT WOULD CAUSE AN INFINITE LOOP
     if(fa[u] == -) return u;
     int tem = father(fa[u]);
     belong[u] = belong[tem];
     fa[u] = tem;
     return tem;
 }

 void solve(){
     if(op == 'D'){
         if(belong[u] == - && belong[v] == -){
             //this is the simplest case
             belong[u] = k++;
             belong[v] = k++;
             nex[u] = v;
             nex[v] = u;
             return;
         }
         if(belong[u] != - && belong[v] == -){
             //notice that v is never mentioned, but u is already processed
             //since is u is vistied, u got its partner
             int fu = father(u), fu1 = father(nex[u]);
             //draw a line from v to u1
             fa[v] = fu1;
             nex[v] = fu;
             belong[v] = belong[fu1];
             return;
             //match a virtue partner for node u
         }
         if(belong[u] == - && belong[v] != -){
             int fv = father(v), fv1 = father(nex[v]);
             fa[u] = fv1;
             nex[u] = fv;
             belong[u] = belong[fv1];
             return;
         }
         if(belong[u] != - && belong[v] != -){
             //notice that both u and v is already visited
             int fu = father(u), fu1 = father(nex[u]);
             int fv = father(v), fv1 = father(nex[v]);
             int bu = belong[fu] >> ;
             int bv = belong[fv] >> ;
             if(bu > bv){
                 // v is relatively primitive
                 fa[fu] = fv1;
                 fa[fu1] = fv;
                 return;
             }
             if(bu < bv){
                 fa[fv] = fu1;
                 fa[fv1] = fu;
                 return;
             }
         }
     }
     if(op == 'A'){
         int fu = father(u);
         int fv = father(v);
         if(belong[fu] == - || belong[fv] == -) puts(str1);
         else if(belong[fu] == belong[fv]) puts(str3);
         else if(belong[fu] == ( ^ belong[fv])) puts(str2);
         else puts(str1);
     }
 }

 int main(){
     freopen("in.txt", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--){
         scanf("%d%d", &n, &m);
         memset(belong, -, sizeof belong);
         memset(fa, -, sizeof fa);
         k = ;
         for(int i = ; i < m; i++){
             scanf(" %c%d%d", &op, &u, &v);
             solve();
         }
     }
 }