CF 444A(DZY Loves Physics-诱导子图的密度)

A. DZY Loves Physics

time limit per test

memory limit per test

input

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.

G, now he wants to find a connected induced subgraph G' of the graph, such that the density of G'

G'(V', E') of a graph G(V, E)

• 

• ;
• edge



• if and only if



• , and edge



• ;
• G' is the same as the value of the corresponding edge inG, 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

n (1 ≤ n ≤ 500), 

. Integer n represents the number of nodes of the graph G, m

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.

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

10 - 9.

Sample test(s)

input

1 01

output

0.000000000000000

input

2 11 2 1 2 1

output

3.000000000000000

input

5 613 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

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;
}