The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 9641		Accepted: 4008

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.

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.

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.

3 3

1 3 2 3 3 1

2 1

1 2

0

1 3

2

定义：点v是汇点须满足 --- 对图中任意点u，若v可以到达u则必有u到v的路径；若v不可以到达u，则u到v的路径可有可无。

题意：在n个点m条边的有向图里面，问有多少个点是汇点。

分析：首先若SCC里面有一个点不是汇点，那么它们全不是汇点，反之也如此。这也就意味着一个SCC里面的点要么全是，要么全不是。在求出SCC并缩点后，任一个编号为A的SCC若存在指向编号为B的SCC的边，那么它里面所有点必不是汇点（因为编号为B的SCC不可能存在指向编号为A的SCC的边）。若编号为A的SCC没有到达其他SCC的路径，那么该SCC里面所有点必是汇点。因此判断的关键在于SCC的出度是否为0.

思路：先用tarjan求出所有SCC，然后缩点后找出所有出度为0的SCC，并用数字存储点，升序排列后输出。

#include<stdio.h>

#include<string.h>

#include<vector>

#include<stack>

#include<algorithm>

#define MAX 21000

#define INF 0x3f3f3f

using namespace std;

int cost[MAX];

int low[MAX],dfn[MAX];

int head[MAX],instack[MAX];

int ans,n,m;

int sccno[MAX],clock;//sccno用来记录当前点属于哪个scc，

int scccnt;//记录总共有多少个scc

stack<int>s;

vector<int>newmap[MAX];//scc缩点之后储存新图

vector<int>scc[MAX];//用来记录scc中的点

int out[MAX];//记录scc的入度

int ant[MAX];

struct node

{

    int beg,end,next;

}edge[MAX];

void init()

{

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

    ans=0;

}

void add(int u,int v)

{

    edge[ans].beg=u;

    edge[ans].end=v;

    edge[ans].next=head[u];

    head[u]=ans++;

}

void getmap()

{

    int i,j,a,b;

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

    {

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

        add(a,b);

    }

}

void tarjan(int u)

{

    int v,i,j;

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

    s.push(u);

    instack[u]=1;

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

    {

        v=edge[i].end;

        if(!dfn[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])

    {

        scccnt++;

        scc[scccnt].clear();//？？

        while(1)

        {

            v=s.top();

            s.pop();

            instack[v]=0;

            sccno[v]=scccnt;

            scc[scccnt].push_back(v);

            if(v==u)

            break;

        }

    }

}

void find(int l,int r)

{

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

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

    memset(sccno,0,sizeof(sccno));

    memset(instack,0,sizeof(instack));

    clock=scccnt=0;

    for(int i=l;i<=r;i++)

    {

        if(!dfn[i])

            tarjan(i);

    }

}

void suodian()

{

    int i,j;

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

    {

        newmap[i].clear();

       out[i]=0;

    }

    for(i=0;i<ans;i++)//遍历所有的边

    {

        int u=sccno[edge[i].beg];//当前边的起点

        int v=sccno[edge[i].end];//当前边的终点

        if(u!=v)//因为sccno中记录的是当前点属于哪个scc，所以u！=v证明不在同一个scc但是由一条边相连，

        {       //证明这两个scc联通

            newmap[u].push_back(v);//将scc中的点储存下来  ？？

            out[u]++;//两个scc联通 则入度加一，

        }

    }

}

void solve()

{

	int i,j,k=0;

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

	{

		if(out[i]==0)

		{

			for(j=0;j<scc[i].size();j++)

			    ant[k++]=scc[i][j];

		}

	}

	sort(ant,ant+k);

	for(i=0;i<k;i++)

	{

        if(i<k-1)

		printf("%d ",ant[i]);

        else

        printf("%d",ant[i]);

	}

	printf("\n");

}

int main()

{

    int j,i;

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

    {

    	scanf("%d",&m);

        init();

        getmap();

        find(1,n);

        suodian();

        solve();

    }

    return 0;

}