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 Fi (1 ≤ Fi ≤ 1000) for each landmark i.
The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 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: Fi
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti
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：每点权val跟边权w，一条路径上的优乐值为点权和/边权和。求牛从一点出发到最后回到原点的最大优乐值
分析：二分参数ans，ans * w1 + ans * w2 + ... + ans * wi <= val1 + val2 + ... + vali;用spfa判断给出的ans能否构成
负环。有说明ans小了，没有说明ans大了。
代码：
#include <stdio.h>
#include <stdlib.h>
#include <queue>
#include <vector>
#define maxn 1005
#define inf 100000000
#define exp 0.00001
using namespace std;
int visit[maxn], val[maxn];
int cnt[maxn];
double dis[maxn];
struct T
{
int v, w;
}e;
vector <T> g[maxn];
void set(int n)
{
for (int i = 1; i <= n; i++)
{
g[i].clear();
}
}
int spfa(int n, int s, double ans)
{
int head = 0, tial = 1, i, u, v, w, size;
double dd;
queue<int> q;
dis[s] = 0;
cnt[s] = 1;
q.push(s);
while (!q.empty())
{
u = q.front();
q.pop();
visit[u] = 0;
size = g[u].size();
for (i = 0; i < size; i++)
{
v = g[u][i].v, w = g[u][i].w;
dd = dis[u] + ans * w - val[v];
if (dis[v] > dd)
{
dis[v] = dd;
if (!visit[v])
{
cnt[v]++;
if (cnt[v] > n)
{
return 0;
}
tial = (tial + 1) % maxn;
q.push(v);
visit[v] = 1;
}
}
}
}
return 1;
}
int main()
{
int n, m, i, u, v, w, th, sta;
double l, r, mid;
while (scanf("%d%d", &n, &m) != EOF)
{
set(n);
for (i = 1, r = 0; i <= n; i++)
{
scanf("%d", &val[i]);
r += val[i];
}
th = 0;
T node;
while (m--)
{
scanf("%d%d%d", &u, &v, &w);
node.v = v, node.w = w;
g[u].push_back(node);
}
l = 0;
while (l + exp < r)
{
mid = (l + r) / 2.0;
for (i = 1; i <= n; i++)
{
dis[i] = inf;
visit[i] = 0;
cnt[i] = 0;
}
sta = spfa(n, 1, mid);
if (sta == 0)
{
l = mid + exp;
}
else
{
r = mid - exp;
}
}
printf("%.2lf\n", l);
}
return 0;
}