poj 2553 强连通分支与缩点

思路：将所有强连通分支找出来，并进行缩点，然后找其中所有出度为0的连通分支，就是题目要求的。

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<queue>

#define Maxn 5100

#define Maxm Maxn*100

#define inf 0x7fffffff

using namespace std;

int index[Maxn],vi[Maxn],stack[Maxn],dfn[Maxn],low[Maxn],e,n,lab,top,num,list,mark[Maxn],degree[Maxn],be[Maxn];

struct Edge{

    int from,to,next;

}edge[Maxm];

void addedge(int from,int to)

{

    edge[e].from=from;

    edge[e].to=to;

    edge[e].next=index[from];

    index[from]=e++;

}

void init()

{

    memset(index,-1,sizeof(index));

    memset(degree,0,sizeof(degree));

    memset(dfn,0,sizeof(dfn));

    memset(low,0,sizeof(low));

    memset(vi,0,sizeof(vi));

    memset(mark,0,sizeof(mark));

    memset(be,0,sizeof(be));

    e=lab=top=num=list=0;

}

void Out(int u)//将该连通分量的点进行统计

{

    int i,j;

    list++;

    do{

        i=stack[--top];

        be[i]=list;

        vi[i]=0;

        mark[i]=1;

    }

    while(i!=u);

}

int dfs(int u)

{

    dfn[u]=low[u]=++lab;

    stack[top++]=u;

    vi[u]=1;

    int i,j,temp;

    for(i=index[u];i!=-1;i=edge[i].next)

    {

        temp=edge[i].to;

        if(!dfn[temp])

        {

            dfs(temp);

            low[u]=min(low[u],low[temp]);

        }

        if(vi[temp])

            low[u]=min(low[u],dfn[temp]);

    }

    if(low[u]==dfn[u])//找到连通分量

        Out(u);

    return 0;

}

int solve()

{

    int i,j,temp;

    for(i=1;i<=n;i++)

        if(!dfn[i])

            dfs(i);

    memset(vi,1,sizeof(vi));

    for(i=1;i<=n;i++)

    {

        for(j=index[i];j!=-1;j=edge[j].next)

        {

            temp=edge[j].to;

            if(be[i]!=be[temp])

            {

                vi[be[i]]=0;

            }

        }

    }

    for(i=1;i<=n;i++)

    {

        if(mark[i])

        {

        if(vi[be[i]])

        {

            printf("%d",i);

            break;

        }

        }

    }

    for(i++;i<=n;i++)

    {

        if(mark[i])

        {

            if(vi[be[i]])

            printf(" %d",i);

        }

    }

    printf("\n");

    return 0;

}

int main()

{

    int m,i,j,a,b;

    while(scanf("%d",&n)!=EOF,n)

    {

        scanf("%d",&m);

        init();

        for(i=1;i<=m;i++)

        {

            scanf("%d%d",&a,&b);

            addedge(a,b);

        }

        solve();

    }

    return 0;

}