POJ 3621 Sightseeing Cows 【01分数规划+spfa判正环】

题目链接：http://poj.org/problem?id=3621

Sightseeing Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11526		Accepted: 3930

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

题目概括：

有 L 个 landmarks， P条 cow path（有向边），每个点可获得娱乐值 Fi ，不过每条边需要花费时间 Ti，我们要求的是选走任意几个点（路径要构成一个环）单位时间获得最大的娱乐值即 求 ΣFi / ΣTi 的最大值。

解题思路：

要从N中取K，并且求 ΣFi / ΣTi 的最大值，很明显用二分+01分数规划

令 ΣFi / ΣTi >= X, 则 ΣFi - ΣTi*X >= 0， 也就转换为了求这个有向图是否存在正环，我们直到SPFA可以轻松通过dfs判断点的访问次数来判断是否有负环，我们只需要把SPFA的求最短路的判断条件换成求最长路的判断条件即可以判断是否存在正环了。

AC code：

///POJ 3621 01分数规划+SPFA判断正环
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
#include <vector>
#include <math.h>
#include <limits>
#include <set>
#include <map>
#define INF ox3f3f3f3f
using namespace std;

const int MAXN = 505050;
int v[MAXN];       ///点权
int fst[MAXN], vb[MAXN], vc[MAXN], nxt[MAXN];
bool vis[MAXN], fh;                     ///记录访问点的次数
double dist[MAXN];    ///用于判断正环
int N, M, cnt;

void add(int a, int b, int c)   ///静态邻接表
{
    ++cnt;
    nxt[cnt] = fst[a];
    fst[a] = cnt;
    vb[cnt] = b;
    vc[cnt] = c;         ///边权
}

void spfa_dfs(int p, double x)
{
    vis[p] = true;
    for(int e = fst[p]; e; e = nxt[e])
    {
        double C = v[vb[e]]-x*vc[e];
        if(dist[vb[e]] >= C+dist[p]) continue;   ///与spfa判断负环恰好相反，只取大的
        if(vis[vb[e]])
        {
            fh = 1; return;
        }
        dist[vb[e]] = C+dist[p];
        spfa_dfs(vb[e], x);
        if(fh) return;
    }
    vis[p] = 0;
}

bool ok(double x)
{
    for(int i = 1; i <= N; i++)
    {
        vis[i] = dist[i] = 0;
    }
    fh = 0;
    for(int i = 1; i <= N; i++)
    {
        spfa_dfs(i, x);
        if(fh) return true;   ///有正环
    }
    return false;
}
int main()
{
    scanf("%d%d", &N, &M);
    for(int i = 1; i <= N; i++)
        scanf("%d", &v[i]);
    for(int i = 1; i <= M; i++)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }
    double l = 0, r = 1000000000;
    while(r-l>1e-6)
    {
        double mid = (l+r)/2.0;
        if(ok(mid)) l = mid;
        else r = mid;
    }
    printf("%.2lf
", l);

    return 0;
}