• CodeForces-916C-Jamie and Interesting Graph


    链接:

    https://vjudge.net/problem/CodeForces-916C

    题意:

    Jamie has recently found undirected weighted graphs with the following properties very interesting:

    The graph is connected and contains exactly n vertices and m edges.
    All edge weights are integers and are in range [1, 109] inclusive.
    The length of shortest path from 1 to n is a prime number.
    The sum of edges' weights in the minimum spanning tree (MST) of the graph is a prime number.
    The graph contains no loops or multi-edges.
    If you are not familiar with some terms from the statement you can find definitions of them in notes section.

    Help Jamie construct any graph with given number of vertices and edges that is interesting!

    思路:

    构造,由1连其他点,权为2,在选1-2补上一个值,为mst权值和减去其他权为2的边的值.
    多余的边补1e9即可.

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int MAXN = 1e5+10;
    pair<pair<int, int>, int > pa[MAXN];
    
    //2*(n-1)
    bool Solve(int x)
    {
        for (int i = 2;i*i <= x;i++)
            if (x%i == 0)
                return true;
        return false;
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n, m;
        cin >> n >> m;
        if (n == 2)
        {
            cout << "2 2" << endl;
            cout << "1 2 2" << endl;
            return 0;
        }
        for (int i = 1;i <= n-1;i++)
            pa[i].first.first = 1, pa[i].first.second = i+1, pa[i].second = 2;
        int v = 2*(n-2)+1;
        while (Solve(v))
            v++;
        pa[1].second = v-2*(n-2);
        int cnt = n;
        for (int i = 2;i <= n;i++)
        {
            if (cnt > m)
                break;
            for (int j = i+1;j <= n;j++)
            {
                pa[cnt].first.first = i;
                pa[cnt].first.second = j;
                pa[cnt].second = 1e9;
                cnt++;
                if (cnt > m)
                    break;
            }
        }
        cout << 2 << ' ' << v << endl;
        for (int i = 1;i <= m;i++)
            cout << pa[i].first.first << ' ' << pa[i].first.second << ' ' << pa[i].second << endl;
    
    
        return 0;
    }
    
  • 相关阅读:
    pytorch nn.Parameters vs nn.Module.register_parameter
    pytorch COCO2017 目标检测 (一)DataLoader
    focal loss 两点理解
    pytorch 目标检测 图像预处理
    C++ 使用copy_if获得数组vector掩膜
    pytorch 网络可视化
    SHELL学习笔记二
    SHELL学习笔记一
    Linux命令笔记一
    VLOOKUP返回#N/A结果
  • 原文地址:https://www.cnblogs.com/YDDDD/p/11392395.html
Copyright © 2020-2023  润新知