POJ2553( 有向图缩点)

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9779		Accepted: 4063

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 Gand 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
题意:给定一幅有向图,若某点所能到达的点也能到达其本身，那么这个点为sink。由小到大输出sink.
思路:有向图缩点得到一棵树,答案为构成叶子（出度为0）结点的连通分量。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=5005;
vector<int> mp[MAXN];
int n,m;
int dfn[MAXN],low[MAXN],time;
int stack[MAXN],top;
bool ins[MAXN];
int belong[MAXN],cnt;
void dfs(int u)
{
    dfn[u]=low[u]=++time;
    stack[top++]=u;
    ins[u]=true;
    for(int i=0;i<mp[u].size();i++)
    {
        int v=mp[u][i];
        if(!dfn[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(ins[v])    low[u]=min(low[u],dfn[v]);
    }
    if(dfn[u]==low[u])
    {
        int v;
        cnt++;
        do{
            v=stack[--top];
            belong[v]=cnt;
            ins[v]=false;
        }while(u!=v);
    }
}
int deg[MAXN];
bool flag[MAXN];
void solve()
{
    /*
    for(int i=1;i<=n;i++)
        printf("%d
",belong[i]);*/
    
    for(int i=1;i<=n;i++)
        for(int j=0;j<mp[i].size();j++)
        {
            int v=mp[i][j];
            if(belong[i]!=belong[v])
            {
                deg[belong[i]]++;
            }
        }
    for(int i=1;i<=cnt;i++)
        if(deg[i]==0)
            flag[i]=true;
    
    for(int i=1;i<=n;i++)
        if(flag[belong[i]])
            printf("%d ",i);
    printf("
");
}
int main()
{
    while(scanf("%d",&n)!=EOF&&n)
    {
        scanf("%d",&m);
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(ins,false,sizeof(ins));
        memset(deg,0,sizeof(deg));
        memset(flag,false,sizeof(flag));
        top=0;
        time=0;
        cnt=0;
        for(int i=1;i<=n;i++)
            mp[i].clear();
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            mp[u].push_back(v);
        }
        for(int i=1;i<=n;i++)
            if(!dfn[i])
                dfs(i);
        solve();
    }
}

下面是kosaraju算法

#include"cstdio"
#include"cstring"
#include"vector"
using namespace std;
const int MAXN=5005;
vector<int> G[MAXN];
vector<int> rG[MAXN];
vector<int> vs;
int V,E;

int cpnt[MAXN];
int vis[MAXN];
void dfs(int u)
{
    vis[u]=1;
    for(int i=0;i<G[u].size();i++)
        if(!vis[G[u][i]])    dfs(G[u][i]);
    vs.push_back(u);
}

void rdfs(int u,int k)
{
    cpnt[u]=k;
    vis[u]=1;
    for(int i=0;i<rG[u].size();i++)
        if(!vis[rG[u][i]])    rdfs(rG[u][i],k);
}

void scc()
{
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=V;i++)
        if(!vis[i])    dfs(i);
    memset(vis,0,sizeof(vis));
    int k=1;
    for(int i=vs.size()-1;i>=0;i--)
        if(!vis[vs[i]])    rdfs(vs[i],k++);
}

int deg[MAXN];
void solve()
{
    scc();
    for(int i=1;i<=V;i++)
    {
        for(int j=0;j<G[i].size();j++)
        {
            int to=G[i][j];
            if(cpnt[i]!=cpnt[to])
            {
                deg[cpnt[i]]++;
            }
        }
    }
    int flag=0;
    for(int i=1;i<=V;i++)
    {
        if(deg[cpnt[i]]==0)
        {
            if(flag==0)
            {
                printf("%d",i);
                flag=1;
            }
            else
            {
                printf(" %d",i);
            }
        }
    }
    printf("
");
}
int main()
{
    while(scanf("%d",&V)!=EOF&&V)
    {
        scanf("%d",&E);
        vs.clear();
        memset(cpnt,0,sizeof(cpnt));
        memset(deg,0,sizeof(deg));
        for(int i=1;i<=V;i++)
        {
            G[i].clear();
            rG[i].clear();
        }
        for(int i=0;i<E;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            rG[v].push_back(u);    
        }
        solve();
    }
        
    return 0;
}