强连通分量 ZQUOJ 21467&&POJ 2553 The Bottom of a Graph

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 x 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) = { vV | w∈V: (vw)  (wv) }. 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

题目大意，给定一个有向图，按顺序输出“自己可达的顶点都可到达自己”的顶点。

解题思路：先求出图的强连通分量，缩点，求出度为0的强连通分量中的顶点，并按按顺序输出。

有关强强连通分量的算法请参考：http://www.byvoid.com/blog/scc-tarjan/（Tarjan算法）

                                          http://www.cnblogs.com/luweiseu/archive/2012/07/14/2591370.html（Tarjan，Kosaraju，Garbow）

                                          http://blog.csdn.net/liguanxing/article/details/5665520（Kosaraju算法）

                                          http://hi.baidu.com/yy17yy/item/8f4142dee291e6ee3cc2cb2f（Garbow算法）

 

AC代码（Tarjan）：

#include<stdio.h>
#include<string.h>
typedef struct
{
    int v,next;
}Node;
Node e[50000];
int n,m,index,top,count;
int first[5001],DFN[5001],Low[5001],ver[5001];
int belong[5001],stack[5001],instack[5001],val[5001],degree[5001];
void Tarjan(int i)
{
    int k,j,v;
    DFN[i]=Low[i]=index++;
    instack[i]=true;
    stack[top++]=i;
    for(j=first[i];j!=-1;j=e[j].next)
    {
        k=e[j].v;
        if(!DFN[k])
        {
            Tarjan(k);
            if(Low[k]<Low[i])
                Low[i]=Low[k];
        }
        else if(instack[k]&&DFN[k]<Low[i])
            Low[i]=DFN[k];
    }
    if(Low[i]==DFN[i])
    {
        count++;
        do{
            v=stack[--top];
            instack[v]=false;
            belong[v]=count;
            val[count]++;
        }while(i!=v);
    }
}
int main()
{
    int g,i,j,u,v;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        g=0; index=0; top=0; count=0;
        memset(first,-1,sizeof(first));
        memset(DFN,0,sizeof(DFN));
        memset(Low,0,sizeof(Low));
        memset(degree,0,sizeof(degree));
        for(i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            e[g].v=v;
            e[g].next=first[u];
            first[u]=g++;
        }
        for(i=1;i<=n;i++)
            if(!DFN[i])
                Tarjan(i);
        for(i=1;i<=n;i++)
            for(j=first[i];j!=-1;j=e[j].next)
            {
                v=e[j].v;
                if(belong[i]!=belong[v])
                    degree[belong[i]]++;
            }
        for(i=1;i<=n;i++)
        {
            if(!degree[belong[i]])
                printf("%d ",i);
        }
        printf("\n");
    }
    return 0;
}

AC代码（Kosaraju）：

#include<stdio.h>
#include<string.h>
typedef struct
{
    int u,v,next1,next2;
}Node;
Node e[50000];
int n,m,index;
int first1[5001],first2[5001],vis[5001],belong[5001],num[5001],degree[5001];
void dfs1(int i)
{
    int j,k;
    vis[i]=1;
    for(j=first1[i];j!=-1;j=e[j].next1)
    {
        k=e[j].v;
        if(!vis[k])
            dfs1(k);
    }
    num[index++]=i;
}
void dfs2(int i)
{
    int j,k;
    vis[i]=1;
    belong[i]=index;
    for(j=first2[i];j!=-1;j=e[j].next2)
    {
        k=e[j].u;
        if(!vis[k])
            dfs2(k);
    }
}
void kosaraju()
{
    int i;
    index=1;
    memset(vis,0,sizeof(vis));
    for(i=1;i<=n;i++)
        if(!vis[i])
            dfs1(i);
    index=1;
    memset(vis,0,sizeof(vis));
    for(i=n;i>=1;i--)
        if(!vis[num[i]])
        {
            dfs2(num[i]);
            index++;
        }
}
int main()
{
    int g,i,j,u,v;
    while(scanf("%d",&n)&&n)
    {
        g=0;
        memset(first1,-1,sizeof(first1));
        memset(first2,-1,sizeof(first2));
        memset(degree,0,sizeof(degree));
        scanf("%d",&m);
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            e[g].v=v;
            e[g].next1=first1[u];
            first1[u]=g;
            e[g].u=u;
            e[g].next2=first2[v];
            first2[v]=g++;
        }
        kosaraju();
        for(i=1;i<=n;i++)
            for(j=first1[i];j!=-1;j=e[j].next1)
            {
                v=e[j].v;
                if(belong[i]!=belong[v])
                    degree[belong[i]]++;
            }
        for(i=1;i<=n;i++)
            if(!degree[belong[i]])
                printf("%d ",i);
        printf("\n");
    }
    return 0;
}

 AC代码（Garbow)：

#include<stdio.h>
#include<string.h>
typedef struct
{
    int v,next;
}Node;
Node e[50000];
int n,m,index,top1,top2,count;
int first[5001],low[5001],stack1[5001],stack2[5001];
int belong[5001],degree[5001];
void Garbow(int i)
{
    int j,k;
    stack1[++top1]=stack2[++top2]=i;
    low[i]=index++;
    for(j=first[i];j!=-1;j=e[j].next)
    {
        k=e[j].v;
        if(!low[k])
            Garbow(k);
        else if(!belong[k])
        {
            while(low[stack2[top2]]>low[k])
                top2--;
        }
    }
    if(stack2[top2]==i)
    {
        top2--;
        count++;
        do{
            k=stack1[top1--];
            belong[k]=count;
        }while(k!=i);
    }
}
int main()
{
    int g,i,j,u,v;
    while(scanf("%d",&n)&&n)
    {
        scanf("%d",&m);
        g=0; top1=top2=-1; index=1; count=0;
        memset(first,-1,sizeof(first));
        memset(low,0,sizeof(low));
        memset(belong,0,sizeof(belong));
        memset(degree,0,sizeof(degree));
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&u,&v);
            e[g].v=v;
            e[g].next=first[u];
            first[u]=g++;
        }
        for(i=1;i<=n;i++)
            if(!low[i])
                Garbow(i);
        for(i=1;i<=n;i++)
            for(j=first[i];j!=-1;j=e[j].next)
            {
                v=e[j].v;
                if(belong[i]!=belong[v])
                    degree[belong[i]]++;
            }
        for(i=1;i<=n;i++)
            if(!degree[belong[i]])
                printf("%d ",i);
        printf("\n");
    }
    return 0;
}