POJ 3621 Sightseeing Cows 最优比率环

Sightseeing Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10569 Accepted: 3623

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 73010105101 2 32 3 23 4 53 5 24 5 55 1 35 2 2

Sample Output

6.00

Source

USACO 2007 December Gold

给了一个有向图，每个点有点权，每条边有边权。求所有环当中，点权和除以边权和最大是多少。

二分答案。

对于一个二分的答案mid,分数规划的分子是点权ai和，分母是点权bi和，则分数规划的函数是sigma(ai-mid*bi).发现环上的点数和边数肯定相等，于是可以在更新边权的同时顺便更新点权，于是只要判断是否存在一个正环就可以。

正向环不是很好求。不过，负环可以用spfa判断。这样，我们就可以先二分答案，每次再通过spfa判负环。判的时候，把点权取负，边权与mid的乘积取整。(把分数规划函数ai-mid*bi取反)

#include <cstdio>#include <iostream>#include <string.h>#include <queue>#include <algorithm>#include <math.h>#include <cmath>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int maxn=1005,maxk=5005,inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const db eps = 1e-5;int head[maxn],t[maxn];bool inq[maxn];db a[maxn],d[maxn];int num;struct Edge {int from,to,pre;db d;};Edge e[maxk*2];void addedge(int from,int to,db d) {e[num].from=from,e[num].to=to,e[num].pre=head[from],e[num].d=d;head[from]=num++;}bool spfa(int n,db mid) {queue<int> q;mem0(t); mem0(inq);for (int i=1;i<=n;i++) d[i]=1e9;inq[1] = 1; q.push(1); d[1] = 0;while (!q.empty()) {int now = q.front(); q.pop();inq[now] = 0;for (int i = head[now]; i != -1; i = e[i].pre) {int to = e[i].to;if (d[now] + mid*e[i].d - a[now] < d[to]) {d[to] = d[now] + mid*e[i].d - a[now];if (!inq[to]) {inq[to] = 1; q.push(to); t[to]++;if (t[to] > n) return true;}}}}return false;}int main() {int n, m, i, j;db l, r, mid, ans;num = 0; memset(head, -1, sizeof(head));scanf("%d%d", &n, &m);for (i = 1; i <= n; i++) scanf("%lf", &a[i]);int x, y;for (i = 1; i <= m; i++) {scanf("%d%d%lf", &x, &y, &r);addedge(x, y, r);}l = 0; r = 1e4;while (true) {mid = (l + r) / 2.0;if (spfa(n,mid)) l = mid; else r = mid;if (fabs(r - l) < eps) break;}printf("%.2lf\n", mid);return 0;}