【缩点+拓扑判链】POJ2762 Going from u to v or from v to u?

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

 

Solution

缩点，满足题意的充要条件是存在一条完整的链，这个可以用“是否存在唯一拓扑序”解决。

Code

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 const int maxn=1e5+;

 int pre[maxn],low[maxn],clock;
 int scc[maxn],s[maxn],c,cnt;
 int head[maxn],e[maxn],nxt[maxn],k;
 void adde(int u,int v){
     e[++k]=v;nxt[k]=head[u];head[u]=k;
 }
 int _head[maxn],_e[maxn],_nxt[maxn],_k;
 void _adde(int u,int v){
     _e[++_k]=v;_nxt[_k]=_head[u];_head[u]=_k;
 }
 int n,m;

 int r[maxn],vis[maxn];
 int topo(){
     for(int i=;i<=cnt;i++){
         int tot=,u;
         for(int i=;i<=cnt;i++)//偷懒
             if(!r[i]&&!vis[i]) tot++,u=i;
         if(tot>) return ;
         vis[u]=;
         for(int i=_head[u];i;i=_nxt[i])
             r[_e[i]]--;
     }
     return ;
 }

 void dfs(int u){
     pre[u]=low[u]=++clock;
     s[++c]=u;
     for(int i=head[u];i;i=nxt[i]){
         int v=e[i];
         if(!pre[v]){
             dfs(v);
             low[u]=min(low[u],low[v]);
         }
         else if(!scc[v]){
             low[u]=min(low[u],pre[v]);
         }
     }
     if(low[u]==pre[u]){
         cnt++;
         while(c){
             scc[s[c]]=cnt;
             if(s[c--]==u) break;
         }
     }
 }

 void clear(){
     memset(head,,sizeof(head));
     memset(_head,,sizeof(_head));
     memset(pre,,sizeof(pre));
     memset(low,,sizeof(low));
     memset(vis,,sizeof(vis));
     memset(r,,sizeof(r));
     c=cnt=k=_k=;
 }

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         clear();
         scanf("%d%d",&n,&m);
         for(int i=;i<=m;i++){
             int u,v;
             scanf("%d%d",&u,&v);
             adde(u,v);
         }
         for(int i=;i<=n;i++)
             if(!pre[i]) dfs(i);

         for(int i=;i<=n;i++)
             for(int j=head[i];j;j=nxt[j]){
                 int u=scc[i],v=scc[e[j]];
                 if(u==v) continue;
                 else _adde(u,v),r[v]++;
             }

         if(topo()) printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }