POJ 2553 The Bottom of a Graph

The Bottom of a Graph

Time Limit: 3000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2553
64-bit integer IO format: %lld      Java class name: Main

 

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 G and 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, epairs 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

Source

Ulm Local 2003

 

解题：求这样的点，它能到的点，那点也可以到它，注意先后顺序。然后求强联通缩点，出度为0的集合，里面的点即为我们求得那些点

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 6000;
struct arc {
    int to,next;
    arc(int x = 0,int y = -1) {
        to = x;
        next = y;
    }
};
arc e[maxn*100];
int head[maxn],dfn[maxn],low[maxn],belong[maxn],my[maxn];
int tot,n,m,top,scc,idx,out[maxn],ans[maxn];
bool instack[maxn];
void init() {
    for(int i = 0; i < maxn; ++i) {
        dfn[i] = low[i] = belong[i] = 0;
        instack[i] = false;
        head[i] = -1;
        out[i] = 0;
    }
    scc = tot = idx = top = 0;
}
void add(int u,int v){
    e[tot] = arc(v,head[u]);
    head[u] = tot++;
}
void tarjan(int u) {
    dfn[u] = low[u] = ++idx;
    my[top++] = u;
    instack[u] = true;
    for(int i = head[u]; ~i; i = e[i].next) {
        if(!dfn[e[i].to]) {
            tarjan(e[i].to);
            low[u] = min(low[u],low[e[i].to]);
        } else if(instack[e[i].to]) low[u] = min(low[u],dfn[e[i].to]);
    }
    if(dfn[u] == low[u]) {
        scc++;
        int v;
        do {
            v = my[--top];
            instack[v] = false;
            belong[v] = scc;
        } while(v != u);
    }
}
int main() {
    int u,v;
    while(~scanf("%d",&n)&&n) {
        scanf("%d",&m);
        init();
        for(int i = 0; i < m; ++i) {
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for(int i = 1; i <= n; ++i)
            if(!dfn[i]) tarjan(i);
        for(int i = 1; i <= n; ++i)
            for(int j = head[i]; ~j; j = e[j].next)
                if(belong[i] != belong[e[j].to]) out[belong[i]]++;
        int cnt = 0;
        for(int i = 1; i <= n; ++i)
            if(!out[belong[i]]) ans[cnt++] = i;
        if(cnt){
            for(int i = 0; i < cnt; ++i)
                printf("%d%c",ans[i],i + 1 == cnt?'
':' ');
        }else puts("");
    }
    return 0;
}