CF 444A(DZY Loves Physics-诱导子图的密度)

A. DZY Loves Physics

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

证明：必然存在一条边数≤1的最优解

假设存在最优解(G)ans最小边数>1，则点数>2

ans=∑vi/∑c 

由假设知对G的子图,(u+v)/c<ans ,(u+v)<ans*c

∴∑u+∑v<ans*∑c ,(∑u+∑v)/∑c<ans=∑vi/∑c

∴(∑u+∑v)<∑vi 矛盾

结论成立

所以只要判断所有的只取1条边，和不取的情况 O(m)

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iomanip>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (500+10)
#define MAXM (MAXN*MAXN)
#define MAXAi (1e6)
#define MAXCi (1e3)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
typedef long double ld;
int n,m,a[MAXN];
ld ans=0.0;
int main()
{
//	freopen("Physics.in","r",stdin);
	scanf("%d%d",&n,&m);
	For(i,n) scanf("%d",&a[i]);
	For(i,m)
	{
		int u,v;
		double c;
		scanf("%d%d%lf",&u,&v,&c);
		ans=max(ans,(ld)(a[u]+a[v])/c);
	}
	cout<<setiosflags(ios::fixed)<<setprecision(100)<<ans;
	
	return 0;
}