tzoj1769: The Bottom of a Graph

原题链接：http://www.tzcoder.cn/acmhome/problemdetail.do?method=showdetail&id=1769

描述

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.

输入

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.

输出

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.

题解：tarjan缩点，在找出出度为0的点。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 20010;//点数
const int MAXM = 50010;//边数
struct Edge{
    int to,next;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong 数组的值是
int Index,top;
int scc;//强连通分量的个数
bool Instack[MAXN];
int num[MAXN];//各个强连通分量包含点的个数，数组编号 1 ～ scc
int du[MAXN];
void addedge(int u,int v){
    edge[tot].to = v;edge[tot].next = head[u];head[u] = tot++;
}
void Tarjan(int u){
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u];i !=-1;i = edge[i].next){
        v = edge[i].to;
        if( !DFN[v] ){
            Tarjan(v);
            if( Low[u] > Low[v] )Low[u] = Low[v];
        }
        else if(Instack[v] && Low[u] > DFN[v])
        Low[u] = DFN[v];
    }
    if(Low[u] == DFN[u]){
        scc++;
        do{
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        }while( v != u);
    }
}
void solve(int N){
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index = scc = top = 0;
    for(int i = 1;i <= N;i++)
        if(!DFN[i])
            Tarjan(i);
}
void init(){
    tot = 0;
    memset(head,-1,sizeof(head));
    memset(edge,0,sizeof(edge));
    memset(du,0,sizeof(du));
}
int main()
{
    int n,m;
    while(scanf("%d",&n))
    {
        if(n==0)break;
        init();
        scanf("%d",&m);
        int u,v;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        solve(n);
        for(int i=1;i<=n;i++)
            for(int j = head[i];j!= -1;j = edge[j].next)
            {
                v=edge[j].to;
                if(Belong[i]!=Belong[v])
                    du[Belong[i]]++;
            }
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            if(du[Belong[i]]==0)
            {
                if(!flag)
                {
                    printf("%d",i);
                    flag=1;
                }
                else printf(" %d",i);
            }
        }
        printf("\n");
    }
    return 0;
}