Poj 3621 Sightseeing Cows 01分数规划

题目链接：http://poj.org/problem?id=3621

Sightseeing Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11523		Accepted: 3928

Description

Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.

Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.

While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values Fi (1 ≤ Fi ≤ 1000) for each landmark i.

The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 1000) to traverse.

In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.

Help the cows find the maximum fun value per unit time that they can achieve.

Input

* Line 1: Two space-separated integers: L and P
* Lines 2..L+1: Line i+1 contains a single one integer: Fi
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti

Output

* Line 1: A single number given to two decimal places (do not perform explicit rounding), the maximum possible average fun per unit time, or 0 if the cows cannot plan any trip at all in accordance with the above rules.

Sample Input

5 7
30
10
10
5
10
1 2 3
2 3 2
3 4 5
3 5 2
4 5 5
5 1 3
5 2 2

Sample Output

6.00

Source

题意：给出一个有向图，每个点都有一个点权，每条有向边也都有一个边权，要求出一个环使得环中点权之和与边权之和的比值最大。

1. //我们想求的是这样一个函数 f = max{ sigma(v[i]*x[i])/sigmax(e[i]*x[i]) } x[i]属于{0，1}

2. //那么设 g(y) = max{sigma(v[i]*x[i])-y*sigmax(e[i]*x[i]}

3. //此函数有两个性质 1.非严格单调递减 2.仅当y为最优解y*的时候，g(y)=0;

4. //当g(y)=0时，y=y*；

5. //当g(y)>0时, y<y*;

6. //当g(y)<0时, y>y*;

7. //所以只要图中存在正环，则g(y)>0  同理如果边反过来储存 如果存在负环 g(y)>0 spfa判断负环即可

#include <iostream>  
#include <stdlib.h>  
#include <stdio.h>  
#include <string.h> 
#include <algorithm>   
#include <math.h>  
#include <cmath>  
#include <vector>  
#include <stack>  
#include <queue> 
using namespace std;
const double eps = 1e-6;
int n,m,cnt;
struct node{
	int u;
	int v;
	int valu;
	int next;
}no[10005];
int a[1005];
int head[1050];
double dist[1050];
int vis[1050];//是否访问过 
int Vis[1050];//访问次数 
void add(int u,int v,int valu)
{
	no[cnt].u=u;
	no[cnt].v=v;
	no[cnt].valu=valu;
	no[cnt].next=head[u];
	head[u]=cnt++;
}
int check(double mid)
{
	memset(vis,0,sizeof(vis));
	memset(Vis,0,sizeof(Vis));
	for(int i=2;i<=n;i++)
		dist[i]=1e15;
	dist[1]=0;
	queue<int>q;
	q.push(1);
	vis[1]=1;
	Vis[1]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=head[u];i!=-1;i=no[i].next)
		{
			int v=no[i].v;
			if(dist[u] - a[u] + mid*no[i].valu < dist[v])
			{
				dist[v] = dist[u] - a[u] + mid*no[i].valu;
				if(vis[v]==0)
				{
					q.push(v);
					Vis[v]++;
					vis[v]=1;
					if(Vis[v]>n)
						return 1;
				}
			}
		}
	} 
	return 0;
}
int main()
{
	cnt=0;
	memset(head,-1,sizeof(head));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for(int i=1;i<=m;i++)
	{
		int u,v,valu;
		scanf("%d%d%d",&u,&v,&valu);
		add(u,v,valu);//单行路 
	}
	double l=0;
	double r=10000;
	while(r-l>eps)
	{
		double mid=(l+r)/2;
		if(check(mid))
			l=mid;
		else
			r=mid;
	}
	printf("%.2lf\n",l);
	return 0;
}