【POJ 3621】Sightseeing Cows

Sightseeing Cows

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 F_i (1 ≤ F_i ≤
1000) for each landmark i.

The cows also know about the cowpaths. Cowpath i connects landmark L_1i to L_2i (in the direction L_1i -> L_2i ) and requires time T_i (1
≤ T_i ≤ 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: F_i

* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L_1i , L_2i , and T_i

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

USACO 2007 December Gold

01分数规划。

求最优比例环。

遇到求sigma(ai)/sigma(bi)就要想到01分数规划了。

设正解为ans。

二分一个答案k,如果k比ans要大那么

即说明当图中没有负环的时候应该减小二分的k值；

有负环的时候应该增大k值。

判断负环的方法与 【BZOJ 1486】一致。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#define eps 1e-4
using namespace std;
double d[1005];
bool f;
int tot=0,v[1005],a[1005],h[1005],n,m;
struct edge
{
	int y,ne,v;
}e[5005];
void Addedge(int x,int y,int v)
{
	tot++;
	e[tot].y=y;
	e[tot].v=v;
	e[tot].ne=h[x];
	h[x]=tot;
}
void dfs(int x,double k)
{
	v[x]=1;
	for (int i=h[x];i;i=e[i].ne)
	{
		int y=e[i].y;
		if (d[y]>d[x]-(double)a[y]+k*(double)e[i].v)
		{
			if (v[y]) {f=true;return;}
			d[y]=d[x]-(double)a[y]+k*(double)e[i].v;
			dfs(y,k);
			if (f) return;
		}
	}
	v[x]=0;
}
bool youfh(double k)
{
	f=false;
	for (int i=1;i<=n;i++)
		d[i]=0.0,v[i]=0;
	for (int i=1;i<=n;i++)
	{
		dfs(i,k);
		if (f) return true;
	}
	return false;
}
void Solve()
{
	double l=0.0,r=2000.0;
	while (r-l>eps)
	{
		double m=(l+r)/(double)2;
		if (youfh(m)) l=m;
		else r=m;
	}
	printf("%.2lf\n",l);
}
int main()
{
        scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	for (int i=1;i<=m;i++)
	{
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		Addedge(x,y,v);
	}
	Solve();
	return 0;
}

感悟：

1.c++就AC，g++就WA

2.对于dfs判断的负环的方法值得借鉴

3.对于01分数规划问题，有两种方法：二分法和Dinkelbach算法。

后者是随便给出一个k，根据通过k求得的值来移动k，使他逼近最优解。

参考：

博客：01分数规划

百度文库：01分数规划