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\n",ans);
}