poj 2553 The Bottom of a Graph

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题意：

给一个定义bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}，意思就是在一个有向图中存在一种点v1，这种点满足:v1可达的点集V中所有点，都有一条路径回到v1。先要找出所有的点，按字典序输出。

思路：

找到出度为0的强联通分量，按字典叙输出其中的点。

#include<iostream>using namespace std;const int MAXN =10001;int DFN[MAXN];int LOW[MAXN];int instack[MAXN];int Stap[MAXN];int Belong[MAXN];int num[MAXN];int count[MAXN];struct edge{int v,next;}vetex[50001];int head[MAXN];int Stop,Bcnt,Dindex,N,k;void add(int a,int b){vetex[k].v = b;vetex[k].next = head[a];head[a] = k;k++;}void tarjan(int i){int j;DFN[i]=LOW[i]=++Dindex;instack[i]=true;Stap[++Stop]=i;for (int k=head[i];k;k=vetex[k].next){j=vetex[k].v;if (!DFN[j]){tarjan(j);if (LOW[j]<LOW[i])LOW[i]=LOW[j];}else if (instack[j] && DFN[j]<LOW[i])LOW[i]=DFN[j];}if (DFN[i]==LOW[i]){Bcnt++;do{j=Stap[Stop--];instack[j]=false;Belong[j]=Bcnt;}while (j!=i);}}void solve(){int i;Stop=Bcnt=Dindex=0;for (i=1;i<=N;i++)if (!DFN[i])tarjan(i);}int main(){int E;while(cin>>N){if(N==0)break;cin>>E;k=1;memset(count,0,sizeof(count));memset(num,0,sizeof(num));memset(head,false,sizeof(head));memset(DFN,0,sizeof(DFN));for(int i=1;i!=E+1;i++){int a,b;cin>>a>>b;add(a,b);}solve();//cout<<Bcnt<<endl;for(int i=1;i!=N+1;i++){int t = Belong[i];for(int k = head[i];k;k=vetex[k].next)if(t!=Belong[vetex[k].v])num[t]++;if(num[t]!=0)count[t]++;}//for(int i=1;i!=Bcnt+1;i++)//cout<<count[i]<<" ";int flag(0);for(int i=1;i!=N+1;i++){int t = Belong[i];if(count[t]==0){if(flag==0)cout<<i;elsecout<<" "<<i;flag=1;}}cout<<endl;}return 0;}