poj2553 强连通缩点

The Bottom of a Graph

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10114		Accepted: 4184

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,...,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.

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

题意:

n个点，m条边，并且是单向边。求有多少个顶点，满足它能到的点也能够到达它。

思路:

对于强连通中的点，肯定能够互相到达，所以可以强连通缩点，此时只要找到出度为0的点，其连通分量连所有的点就是答案。因为

如果该点的出度不为0，那么肯定有新的该点能够到达的点，由于已经缩点了，不可能出现有强连通的情况，所以出度不为0的点不满足要求。

/*

 * Author:  sweat123

 * Created Time:  2016/6/25 14:32:24

 * File Name: main.cpp

 */

#include<set>

#include<map>

#include<queue>

#include<stack>

#include<cmath>

#include<string>

#include<vector>

#include<cstdio>

#include<time.h>

#include<cstring>

#include<iostream>

#include<algorithm>

#define INF 1<<30

#define MOD 1000000007

#define ll long long

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define pi acos(-1.0)

using namespace std;

const int MAXN = ;

struct node{

    int from;

    int to;

    int next;

}edge[MAXN*];

int pre[MAXN],vis[MAXN],dfn[MAXN],low[MAXN],n,m,ind;

int f[MAXN],siz[MAXN],num,dep,out[MAXN],in[MAXN];

stack<int>s;

vector<int>p[MAXN];

void add(int x,int y){

    edge[ind].from = x;

    edge[ind].to = y;

    edge[ind].next = pre[x];

    pre[x] = ind ++;

}

void dfs(int rt){

    dfn[rt] = low[rt] = ++dep;

    vis[rt] = ;

    s.push(rt);

    for(int i = pre[rt]; i != -; i = edge[i].next){

        int t = edge[i].to;

        if(!dfn[t]){

            dfs(t);

            low[rt] = min(low[rt],low[t]);

        } else if(vis[t]){

            low[rt] = min(low[rt],dfn[t]);

        }

    }

    if(low[rt] == dfn[rt]){

        ++num;

        while(!s.empty()){

            int  tp = s.top();

            s.pop();

            vis[tp] = ;

            f[tp] = num;

            siz[num] ++;

            p[num].push_back(tp);

            if(tp == rt)break;

        }

    }

}

void setcc(){

    num = ;

    dep = ;

    memset(in,,sizeof(in));

    memset(out,,sizeof(out));

    memset(dfn,,sizeof(dfn));

    memset(low,,sizeof(low));

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

        if(!dfn[i]){

            dfs(i);

        }

    }

    memset(pre,-,sizeof(pre));

    int ret = ind;

    for(int i = ; i < ret; i++){

        int x = f[edge[i].from];

        int y = f[edge[i].to];

        if(x == y)continue;

        add(x,y);

        out[x] ++;

        in[y] ++;

    }

    vector<int>ans;

    for(int i = ; i <= num; i++){

        if(!out[i]){

            for(int j = ; j < p[i].size(); j++){

                ans.push_back(p[i][j]);

            }

        }

    }

    sort(ans.begin(),ans.end());

    for(int i = ; i < ans.size(); i++){

        if(i == )printf("%d",ans[i]);

        else printf(" %d",ans[i]);

    }

    printf("\n");

}

int main(){

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

        if(n == )break;

        scanf("%d",&m);

        if(n == ){

            printf("1\n");

            continue;

        }

        ind = ;

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

            p[i].clear();

        }

        memset(pre,-,sizeof(pre));

        while(!s.empty())s.pop();

        memset(f,-,sizeof(f));

        memset(siz,,sizeof(siz));

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

            int x,y;

            scanf("%d%d",&x,&y);

            add(x,y);

        }

        setcc();

    }

    return ;

}