• Codeforces Round #254 (Div. 1) A. DZY Loves Physics 智力题


    A. DZY Loves Physics

    题目连接:

    http://codeforces.com/contest/444/problem/A

    Description

    DZY loves Physics, and he enjoys calculating density.

    Almost everything has density, even a graph. We define the density of a non-directed graph (nodes and edges of the graph have some values) as follows:

    where v is the sum of the values of the nodes, e is the sum of the values of the edges.
    Once DZY got a graph G, now he wants to find a connected induced subgraph G' of the graph, such that the density of G' is as large as possible.

    An induced subgraph G'(V', E') of a graph G(V, E) is a graph that satisfies:

    ;
    edge if and only if , and edge ;
    the value of an edge in G' is the same as the value of the corresponding edge in G, so as the value of a node.
    Help DZY to find the induced subgraph with maximum density. Note that the induced subgraph you choose must be connected.

    Input

    The first line contains two space-separated integers n (1 ≤ n ≤ 500), . Integer n represents the number of nodes of the graph G, m represents the number of edges.

    The second line contains n space-separated integers xi (1 ≤ xi ≤ 106), where xi represents the value of the i-th node. Consider the graph nodes are numbered from 1 to n.

    Each of the next m lines contains three space-separated integers ai, bi, ci (1 ≤ ai < bi ≤ n; 1 ≤ ci ≤ 103), denoting an edge between node ai and bi with value ci. The graph won't contain multiple edges.

    Output

    Output a real number denoting the answer, with an absolute or relative error of at most 10 - 9.

    Sample Input

    1 0
    1

    Sample Output

    0.000000000000000

    题意

    给你一个带边权和带点权的无向图,你需要找到一个最大的子团,使得这个子团的点权和除以边权和最大

    题解:

    最多选择两个点,这个很容易证明

    然后暴力莽一波就好了

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 505;
    vector<pair<int,double> >E[maxn];
    double val[maxn];
    int n,m,cnt;
    double ans;
    int main()
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%lf",&val[i]);
        for(int i=1;i<=m;i++)
        {
            int a,b;double c;
            scanf("%d%d%lf",&a,&b,&c);
            ans=max(ans,(val[a]+val[b])/c);
        }
        printf("%.12f
    ",ans);
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/5468613.html
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