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
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const db  PI = acos(-1.0);
const db  eps = 1e-10;
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]=1;
    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]=0;
        }while(u!=v);
        num++;
    }
}
int main(){
    while (scanf("%d%d",&n,&m)==2&&n){
        memset(head,-1, sizeof(head));
        memset(low,0, sizeof(low));
        memset(dfn,0, sizeof(dfn));
        memset(ins,0, sizeof(ins));
        memset(out,0, sizeof(out));
        cnt=num=id=0;
        for(int i=0;i<m;i++){
            int x,y;
            ci(x),ci(y);
            add(x,y);
        }
        for(int i=1;i<=n;i++)
            if(!dfn[i]) tarjan(i);
        for(int i=1;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=1;
        for(int i=1;i<=n;i++){
            if(!out[beg[i]]){//所在强连通分量出度为0
                if(ok) printf("%d",i),ok=0;
                else printf(" %d",i);
            }
        }
        puts("");
    }
    return 0;
}