POJ2553 汇点个数(强连通分量

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12070		Accepted: 4971

Description

We will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being called vertices (or nodes). Let E be a subset of the Cartesian product V×V, its elements being called edges. Then G=(V,E) is called a directed graph. 
Let n be a positive integer, and let p=(e₁,...,e_n) be a sequence of length n of edges e_i∈E such that e_i=(v_i,v_i+1) for a sequence of vertices (v₁,...,v_n+1). Then p is called a path from vertex v₁ to vertex v_n+1 in G and we say that v_n+1 is reachable from v₁, writing (v₁→v_n+1). 
Here are some new definitions. A node v in a graph G=(V,E) is called a sink, if for every node w in G that is reachable from v, v is also reachable from w. The bottom of a graph is the subset of all nodes that are sinks, i.e., bottom(G)={v∈V|∀w∈V:(v→w)⇒(w→v)}. You have to calculate the bottom of certain graphs.

Input

The input contains several test cases, each of which corresponds to a directed graph G. Each test case starts with an integer number v, denoting the number of vertices of G=(V,E), where the vertices will be identified by the integer numbers in the set V={1,...,v}. You may assume that 1<=v<=5000. That is followed by a non-negative integer e and, thereafter, e pairs of vertex identifiers v₁,w₁,...,v_e,w_e with the meaning that (v_i,w_i)∈E. There are no edges other than specified by these pairs. The last test case is followed by a zero.

Output

For each test case output the bottom of the specified graph on a single line. To this end, print the numbers of all nodes that are sinks in sorted order separated by a single space character. If the bottom is empty, print an empty line.

Sample Input

3 3
1 3 2 3 3 1
2 1
1 2
0

Sample Output

1 3
2

Source

Ulm Local 2003

定义：点v是汇点须满足 --- 对图中任意点u，若v可以到达u则必有u到v的路径；若v不可以到达u，则u到v的路径可有可无。

题意：在n个点m条边的有向图里面，问有多少个点是汇点。

分析：若SCC里有一点不是汇点，那么它们全不是，反之也如此。所以一个SCC里面的点要么全是，要么全不是。在求出SCC并缩点后，任一个编号为A的SCC若存在指向编号为B的SCC的边，那么其所有点必不是汇点（因为编号为B的SCC不可能存在指向编号为A的SCC的边）。若编号为A的SCC没有到达其他SCC的路径，那么该SCC里面所有点必是汇点。因此判断的关键在于SCC的出度是否为0.

思路：先用tarjan求出所有SCC，然后缩点后找出所有出度为0的SCC，并用数字存储点，升序排列后输出。

代码：

 #include"bits/stdc++.h"

 #define db double
 #define ll long long
 #define vl vector<ll>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define rep(i, n) for(int i=0;i<n;i++)
 using namespace std;
 const int N = 1e6 + ;
 const int mod = 1e9 + ;
 const int MOD = ;
 const db  PI = acos(-1.0);
 const db  eps = 1e-;
 const ll  INF = 0x3fffffffffffffff;
 int n,m;
 struct P{int to,nxt;}e[N];
 int head[N];
 bool ins[N];
 int beg[N];
 int low[N],dfn[N];
 int out[N];
 stack<int> s;

 int cnt,num,id;
 void add(int u,int v){
     e[cnt].to=v;
     e[cnt].nxt=head[u];
     head[u]=cnt++;
 }
 void tarjan(int u)
 {
     low[u]=dfn[u]=++id;
     ins[u]=;
     s.push(u);
     for(int i=head[u];~i;i=e[i].nxt){
         int v=e[i].to;
         if(!dfn[v]) tarjan(v),low[u]=min(low[u],low[v]);
         else if(ins[v]) low[u]=min(low[u],dfn[v]);
     }
     if(low[u]==dfn[u]){
         int v;
         do{
             v=s.top();s.pop();
             beg[v]=num;//强连通分量编号
             ins[v]=;
         }while(u!=v);
         num++;
     }
 }
 int main(){
     while (scanf("%d%d",&n,&m)==&&n){
         memset(head,-, sizeof(head));
         memset(low,, sizeof(low));
         memset(dfn,, sizeof(dfn));
         memset(ins,, sizeof(ins));
         memset(out,, sizeof(out));
         cnt=num=id=;
         for(int i=;i<m;i++){
             int x,y;
             ci(x),ci(y);
             add(x,y);
         }
         for(int i=;i<=n;i++)
             if(!dfn[i]) tarjan(i);
         for(int i=;i<=n;i++){
             for(int j=head[i];~j;j=e[j].nxt){
                 int v=e[j].to;
                 if(beg[i]!=beg[v]) out[beg[i]]++;
             }
         }
         bool ok=;
         for(int i=;i<=n;i++){
             if(!out[beg[i]]){//所在强连通分量出度为0
                 if(ok) printf("%d",i),ok=;
                 else printf(" %d",i);
             }
         }
         puts("");
     }
     return ;
 }