【POJ3621】Sightseeing Cows

Sightseeing Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8331		Accepted: 2791

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

题意：给出一个有向图 问求一个回路 使得回路上的点权之和/边权之和最大

01分数规划，简单构造，将点权转移到边权上~因为一个环上的点和边的数量是相等的~

设i,j之间初始边权为w[i][j]，修改后的边权为g[i][j]，则g[i][j]=w[i][j]*mid+val[i]

spfa判负环即可~

其实说起来很简单写的时候好烦。。。。因为G++%lf输出问题WA了好多次。。。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>
 using namespace std;

 const int maxn = ;
 const int maxm = ;
 int a[maxm], b[maxm];
 int head[maxn], next[maxm], to[maxm];
 int q[maxn*], im[maxm];
 int vis[maxn];
 int cnt, st;
 int n, m;
 double l, r, mid;
 double c[maxm], dis[maxn], len[maxm], val[maxn];

 inline void add(int u, int v, double w);
 bool dfs(int u);
 bool spfa();

 int main(){
     while(scanf("%d%d", &n, &m)!=EOF){
         for(int i = ; i <= n; ++i){
             scanf("%lf", &val[i]);
         }

         memset(head, -, sizeof(head));
         cnt = ;
         for(int i = ; i <= m; ++i){
             scanf("%d%d%lf", &a[i], &b[i], &c[i]);
             add(a[i], b[i], c[i]);
         }

         l = 0.0;
         r = 1000.0;
         while(r-l > 1e-){
             mid = (r+l) / 2.0;
             if(spfa()){
                 l = mid;
             }
             else{
                 r = mid;
             }
         }
         printf("%.2f\n", mid);
     }
     return ;
 }

 inline void add(int u, int v, double w){
     to[cnt] = v;
     len[cnt] = w;
     next[cnt] = head[u];
     head[u] = cnt++;
 }

 bool dfs(int u){
     vis[u] = st;
     for(int i = head[u]; ~i; i = next[i]){
         if(dis[to[i]] > dis[u] + len[i]*mid - val[u]){
             dis[to[i]] = dis[u] + len[i]*mid - val[u];
             if(vis[to[i]] == st){
                 return true;
             }
             else if(dfs(to[i])){
                 return true;
             }
         }
     }
     vis[u] = ;
     return false;
 }

 bool spfa(){
     memset(vis, , sizeof(vis));
     for(st = ; st <= n; ++st){
         if(dfs(st)){
             return true;
         }
     }
     return false;
 }