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 test(s)
input
1 0
1
output
0.000000000000000
input
2 1
1 2
1 2 1
output
3.000000000000000
input
5 6
13 56 73 98 17
1 2 56
1 3 29
1 4 42
2 3 95
2 4 88
3 4 63
output
2.965517241379311
Note
In the first sample, you can only choose an empty subgraph, or the subgraph containing only node 1.
In the second sample, choosing the whole graph is optimal.
題意： 給出一個源圖， 要求尋找一個密度（點的值/邊的值）最大的子圖；
#include<stdio.h>
#include<iostream>
using namespace std;
int a[510];
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
double ans=0.0;
while(m--)
{
int f,t,v;
scanf("%d%d%d",&f,&t,&v);
ans=max(ans,double(a[f]+a[t])/v);
}
printf("%.11f\n",ans);
}