Gym-101630C Connections

Connections

思维

真没看出来

由于给的是强连通图，直接保留从 \(1\) 开始深搜，保证 \(1\) 能访问到其他所有点的，然后反向建图，保证所有点能够到达 \(1\)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
const int maxn = 1e5 + 10;
vector<int>gra[2][maxn];
int ans[2][maxn];
int vis[maxn];

void dfs(int now, int pre, int f)
{
    if(ans[f][now]) return;
    ans[f][now] = pre;
    for(int nex : gra[f][now])
        dfs(nex, now, f);
}

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int n, m;
        cin >> n >> m;
        for(int i=0; i<2; i++)
        {
            for(int j=0; j<=n; j++)
            {
                gra[i][j].clear();
                ans[i][j] = 0;
            }
        }
        
        for(int i=0; i<m; i++)
        {
            int a, b;
            cin >> a >> b;
            gra[0][a].push_back(b);
            gra[1][b].push_back(a);
        }
        dfs(1, 1, 1);
        dfs(1, 1, 0);
        int cnt = m - 2 * n;
        int cur = 0;
        for(int i=1; i<=n && cnt; i++)
        {
            for(int j=0; j<gra[0][i].size() && cnt; j++)
            {
                int nex = gra[0][i][j];
                if(ans[0][nex] == i || ans[1][i] == nex) continue;
                if(cur++) cout << " ";
                cout << i << " " << nex << "\n";
                cnt--;
            }
        }
    }
    return 0;
}

标签：now,101630C,int,Gym,dfs,Connections,maxn,ans,include
来源： https://www.cnblogs.com/dgsvygd/p/16611137.html