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,...,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
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 = 5010;
struct node{
int from;
int to;
int next;
}edge[MAXN*10];
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] = 1;
s.push(rt);
for(int i = pre[rt]; i != -1; 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] = 0;
f[tp] = num;
siz[num] ++;
p[num].push_back(tp);
if(tp == rt)break;
}
}
}
void setcc(){
num = 0;
dep = 0;
memset(in,0,sizeof(in));
memset(out,0,sizeof(out));
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
for(int i = 1; i <= n; i++){
if(!dfn[i]){
dfs(i);
}
}
memset(pre,-1,sizeof(pre));
int ret = ind;
for(int i = 0; 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 = 1; i <= num; i++){
if(!out[i]){
for(int j = 0; j < p[i].size(); j++){
ans.push_back(p[i][j]);
}
}
}
sort(ans.begin(),ans.end());
for(int i = 0; i < ans.size(); i++){
if(i == 0)printf("%d",ans[i]);
else printf(" %d",ans[i]);
}
printf("\n");
}
int main(){
while(~scanf("%d",&n)){
if(n == 0)break;
scanf("%d",&m);
if(n == 1){
printf("1\n");
continue;
}
ind = 0;
for(int i = 1; i <= n; i++){
p[i].clear();
}
memset(pre,-1,sizeof(pre));
while(!s.empty())s.pop();
memset(f,-1,sizeof(f));
memset(siz,0,sizeof(siz));
for(int i = 1; i <= m; i++){
int x,y;
scanf("%d%d",&x,&y);
add(x,y);
}
setcc();
}
return 0;
}
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,...,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