Codeforces 444A DZY Loves Physics(贪心，数学)

题目链接

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 subgraphG' of the graph, such that the density of G' is as large as possible.

An induced subgraph G'(V', E') of a graphG(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 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

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

The second line contains n space-separated integersx_i (1 ≤ x_i ≤ 10⁶), wherex_i represents the value of thei-th node. Consider the graph nodes are numbered from1 to n.

Each of the next m lines contains three space-separated integersa_i, b_i, c_i (1 ≤ a_i < b_i ≤ n; 1 ≤ c_i ≤ 10³), denoting an edge between node a_i andb_i with valuec_i. The graph won't contain multiple edges.

Output

Output a real number denoting the answer, with an absolute or relative error of at most10^- 9.

Sample test(s)

Input

1 01

Output

0.000000000000000

Input

2 11 21 2 1

Output

3.000000000000000

Input

5 613 56 73 98 171 2 561 3 291 4 422 3 952 4 883 4 63

Output

2.965517241379311

题意：n个点的无向图。每个点有点权，边有边权。一个图的密度的定义为：

1，顶点只有一个，则密度为0

2，顶点大于1个，密度等于v/e，v为点权和，e为边权和。

求该图的所有子图中，最大的密度值。

题解：可以证明密度最大的子图，一定只有两个点。

假设一个图现在有两个点点权为v1，v2，他们之间相连的边的边权为m1，该图的密度为(v1+v2)/m1。如果增加一个点v3要让该图的密度增加，若v3与v2相连的边的边权为m2。那么只有与v3/m2>(v1+v2)/m1，该图的密度才会增加。但是此时，v2与v3两个点构成的子图的密度为(v2+v3)/m2>(v1+v2+v3)/(m1+m2)。所以密度最大的子图一定只有两个点。

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<string>#include<stack>#include<math.h>#include<vector>#include<set>#include<map>#define nn 110000#define inff 0x3fffffff#define eps 1e-8#define mod 1000000007typedef long long LL;const LL inf64=LL(inff)*inff;using namespace std;int n,m;double a[510];int main(){    int i;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%lf",&a[i]);        }        double ans=0;        int u,v;        double l;        for(i=1;i<=m;i++)        {            scanf("%d%d%lf",&u,&v,&l);            ans=max(ans,(a[u]+a[v])/l);        }        printf("%.10lf\n",ans);    }    return 0;}