【NOIP2015】【Luogu2661】信息传递(有向图最小环)

problem

  • 给定一张n个点,n条边的有向图
  • 求图的最小环,输出大小

solution

kosaraju暴力求出所有强连通分量,取最小值即可。

codes

//kosaraju
#include<iostream>
#include<algorithm>
#include<vector>
#define maxn 200010
using namespace std;
vector<int>G[maxn], rG[maxn];
vector<int>vs, cmp[maxn];
int vis[maxn], book[maxn], cnt;
void dfs(int u){
    if(vis[u])return ;
    vis[u] = 1;
    for(int i = 0; i < G[u].size(); i++)dfs(G[u][i]);
    //for(int i = 0; i < G[u].size(); i++)if(!vis[G[u][i]])dfs(G[u][i]);
    vs.push_back(u);
}
void rdfs(int u){
    //if(book[u])return ;
    book[u] = cnt;
    cmp[cnt].push_back(u);
    for(int i = 0; i < rG[u].size(); i++)if(!book[rG[u][i]])rdfs(rG[u][i]);
    //for(int i = 0; i < rG[u].size(); i++)rdfs(rG[u][i]);
}
int main(){
    int n;   cin>>n;
    for(int i = 1; i <= n; i++){
        int x;  cin>>x;
        G[i].push_back(x);
        rG[x].push_back(i);
    }
    for(int i = 1; i <= n; i++)dfs(i);
    for(int i = n-1; i >= 0; i--){
        if(!book[vs[i]]){
            ++cnt;
            rdfs(vs[i]);
        }
    }
    int ans = 0xffffff;
    for(int i = 1; i <= cnt; i++){
        int si = cmp[i].size();
        if(si != 0 && si != 1){
            ans = min(ans, si);
        }
    }
    cout<<ans<<"\n";
    return 0;
}
posted @ 2018-08-14 23:01  gwj1139177410  阅读(154)  评论(0编辑  收藏  举报
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