P2868 [USACO07DEC]Sightseeing Cows G

每个边都有边权，每个点都有点权，题意即找到一个环，使得经过的点权和与边权和的比值最大，对于一个环，点数和边数一样，我们将所有的点权放在到达该点的边上，记点权为 $f_{i}$ ，边权为 $w_{i}$ ，经过了 $k$ 条边，则要使得 $\frac{\sum_{i = 1}^{k} f_{i}}{\sum_{i = 1}^{k} w_{i}}$ 最大，考虑分数规划。

设当前二分平均值为 $x$ ，则 $x$ 合法的条件为：
$\begin{aligned} \frac{\sum_{i = 1}^{k} f_{i}}{\sum_{i = 1}^{k} w_{i}} \geq x \\ ⟹ & \sum_{i = 1}^{k} f_{i} \geq x \sum_{i = 1}^{k} w_{i} \\ ⟹ & \sum_{i = 1}^{k} f_{i} - x w_{i} \geq 0 \end{aligned}$
即需要找到一个正环，所有点都可以作为起点，所以都放进队列里做 spfa 最长路即可。

查看代码

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
const double eps = 1e-4;
const int N = 1e3 + 10, M = 5e3 + 10;
double d[N];
bool vis[N];
int n, m, cnt[N], f[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool check(double mid)
{
    for (int i = 1; i <= n; i++)
        d[i] = cnt[i] = 0;
    queue <int> q;
    for (int i = 1; i <= n; i++)
    {
        q.push(i);
        vis[i] = true;
    }
    while (!q.empty())
    {
        int t = q.front();
        q.pop();
        vis[t] = false;
        for (int i = hd[t]; ~i; i = nxt[i])
            if (d[t] + f[t]  - mid * wt[i] > d[edg[i]])
            {
                d[edg[i]] = d[t] + f[t] - mid * wt[i];
                cnt[edg[i]] = cnt[t] + 1;
                if (cnt[edg[i]] >= n)
                    return true;
                if (!vis[edg[i]])
                {
                    q.push(edg[i]);
                    vis[edg[i]] = true;
                }
            }
    }
    return false;
}
void add(int x, int y, int z)
{
    nxt[++idx] = hd[x];
    hd[x] = idx;
    edg[idx] = y;
    wt[idx] = z;
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        hd[i] = -1;
    for (int i = 1; i <= n; i++)
        cin >> f[i];
    for (int i = 1, a, b, c; i <= m; i++)
    {
        cin >> a >> b >> c;
        add(a, b, c);
    }
    double l = 0, r = N, mid;
    while (r - l > eps)
    {
        mid = (l + r) / 2;
        if (check(mid))
            l = mid;
        else
            r = mid;
    }
    printf("%.2lf", l);
    return 0;
}