hoj 1520 The Bottom of a Graph // poj 2553 The Bottom of a Graph

/*



找到强连通分量变成缩点后求给出出度为0的所有点

以下用tarjan算法做，第一个是用结构体的邻接链表来做，第二个使用的是vector作为邻接链表



*/

#include <cstdio>

#include <cstring>



const int X = 15002;



int dfn[X],low[X],stack[X],father[X],depth,top,bcnt;

int counter[X],n,m;

bool instack[X];



struct node

{

    int v;

    node *next;

    void fun()

    {

        v = 0;

        next = NULL;

    }

}edge[X],*head[X],*tmp;



void tarjan(int u)

{

    int v;

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

    stack[++top] = u;

    instack[u] = true;



    for(node *p=head[u];p;p=p->next)

    {

        v = p->v;

        if(!low[v])

        {

            tarjan(v);

            if(low[v]<low[u])

                low[u] = low[v];

        }

        else if(instack[v]&&low[u]>dfn[v])

            low[u] = dfn[v];

    }

    if(low[u]==dfn[u])

    {

        bcnt++;

        do

        {

            v = stack[top--];

            instack[v] = false;

            father[v] = bcnt;

        }while(u!=v);

    }

}



void solve()

{

    top = depth = bcnt = 0;

    memset(instack,false,sizeof(instack));

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

    memset(counter,0,sizeof(counter));



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

        if(!low[i])

            tarjan(i);



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

        for(node *p=head[i];p;p=p->next)

            if(father[i]!=father[p->v])

                counter[father[i]]++;

    bool flag = false;

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

        if(!counter[father[i]])

        {

            if(flag)

                printf(" ");

            else

                flag = true;

            printf("%d",i);

        }

    printf("\n");

}



int main()

{

    freopen("sum.in","r",stdin);

    freopen("sum.out","w",stdout);

    int u,v;

    while(scanf("%d",&n),n)

    {

        scanf("%d",&m);

        memset(head,NULL,sizeof(head));



        for(int i=0;i<m;i++)

            edge[i].fun();

        tmp = edge;

        for(int i=0;i<m;i++)

        {

            scanf("%d%d",&u,&v);

            tmp->next = head[u];

            tmp->v = v;

            head[u] = tmp++;

        }

        solve();

    }

    return 0;

}

/*



vector作为邻接链表



*/

#include <vector>

#include <cstdio>

#include <cstring>



using namespace std;



const int X = 15005;



int dfn[X],low[X],father[X],stack[X],bcnt,top,depth;

int counter[X],n,m;

bool instack[X];

vector<int> adj[X];



void tarjan(int u)

{

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

    instack[u] = true;

    stack[++top] = u;

    int v,len = adj[u].size();

    for(int i=0;i<len;i++)

    {

        v = adj[u][i];

        if(!low[v])

        {

            tarjan(v);

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

        }

        else if(instack[v])

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

    }

    if(low[u]==dfn[u])

    {

        ++bcnt;

        do

        {

            v = stack[top--];

            instack[v] = false;

            father[v] = bcnt;

        }while(u!=v);

    }

}



void solve()

{

    depth = top = bcnt = 0;

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

    memset(instack,false,sizeof(instack));

    memset(counter,0,sizeof(counter));



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

        if(!low[i])

            tarjan(i);



    int len;

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

    {

        len = adj[i].size();

        for(int j=0;j<len;j++)

            if(father[i]!=father[adj[i][j]])

                counter[father[i]]++;

    }

    bool flag = false;

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

        if(!counter[father[i]])

        {

            if(flag)

                printf(" ");

            else

                flag = true;

            printf("%d",i);

        }

    printf("\n");

}



int main()

{

    freopen("sum.in","r",stdin);

    freopen("sum.out","w",stdout);

    int u,v;

    while(scanf("%d",&n),n)

    {

        scanf("%d",&m);

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

            adj[i].clear();

        for(int i=0;i<m;i++)

        {

            scanf("%d%d",&u,&v);

            adj[u].push_back(v);

        }

        solve();

    }



    return 0;

}