POJ 3621 Sightseeing Cows（最优比例环+SPFA检测）

Posted 2020-09-22 Blackops

Sightseeing Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10306		Accepted: 3519

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

 

题目链接：POJ 3621

最优比例环的题，也是用01分数规划写，目的是找到一个环，使得该环上${\Sigma w_v \over \Sigma w_e}$最大，那么我们设比例为r，当${\Sigma w_v \over \Sigma w_e}>r$时说明可以找到更大的r‘作为答案，由这个式子有可以得到：$\Sigma w_v - r * \Sigma w_e>0$，即存在左边的式子结果>0即可，若要找存在一个数大于0，那么肯定找这个数可能的最大值，那显然把边权重新赋值为$w_{vi}-r*w_{e}$，然后这式子跟SPFA找正环有什么关系？可以发现$\Sigma w_v - r * \Sigma w_e$这个式子代表了这个环上用$w_{vi}-r*w_{ei}$作为新边权的所有边权之和，如果这个和大于0，那显然这个环上存在正环，即有不存在最长路，那每一次SPFA找最长路看看是否存在即可，当然也把式子加个负号，然后用负环检测也可以做

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1010;
const int M = 5010;
const double eps = 1e-6;
struct edge
{
    int to, nxt;
    double w;
    edge() {}
    edge(int _to, int _nxt, double _w): to(_to), nxt(_nxt), w(_w) {}
};
edge E[M];
int head[N], tot;
int vis[N], cnt[N];
double d[N];
int arr[N];

void init()
{
    CLR(head, -1);
    tot = 0;
}
inline void add(int s, int t, double w)
{
    E[tot] = edge(t, head[s], w);
    head[s] = tot++;
}
int spfa(int n, double g)
{
    queue<int>Q;
    for (int i = 1; i <= n; ++i)
    {
        d[i] = 0;
        vis[i] = 1;
        cnt[i] = 1;
        Q.push(i);
    }
    while (!Q.empty())//找正环
    {
        int u = Q.front();
        Q.pop();
        vis[u] = 0;
        for (int i = head[u]; ~i; i = E[i].nxt)
        {
            int v = E[i].to;
            double w = arr[u] - g * E[i].w;
            if (d[v] < d[u] + w)
            {
                d[v] = d[u] + w;
                if (!vis[v])
                {
                    vis[v] = 1;
                    Q.push(v);
                    if (++cnt[v] > n)
                        return 1;
                }
            }
        }
    }
    return 0;
}
int main(void)
{
    int n, m, a, b, i;
    double w;
    while (~scanf("%d%d", &n, &m))
    {
        init();
        for (i = 1; i <= n; ++i)
            scanf("%d", &arr[i]);
        for (i = 1; i <= m; ++i)
        {
            scanf("%d%d%lf", &a, &b, &w);
            add(a, b, w);
        }
        double L = 0, R = 1000;
        double ans = 0;
        while (fabs(R - L) >= eps)
        {
            double mid = (L + R) / 2.0;
            if (spfa(n, mid))
            {
                L = mid;
                ans = mid;
            }
            else
                R = mid;
        }
        printf("%.2f\n", ans);
    }
    return 0;
}