poj 2553 The Bottom of a Graph【强连通分量求汇点个数】

The Bottom of a Graph
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 9641   Accepted: 4008

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,...,en) be a sequence of length n of edges ei∈E such that ei=(vi,vi+1) for a sequence of vertices (v1,...,vn+1). Then p is called a path from vertex v1 to vertex vn+1 in Gand we say that vn+1 is reachable from v1, writing (v1→vn+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,...,ve,we with the meaning that (vi,wi)∈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


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。

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#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;
}

　　

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