[ABC324F] Beautiful Path - 洛谷

思路

首先看到这个形式很容易想到 01 分数规划，即去二分答案，然后就是转化成 是否存在一个路径使得 sigma b - mid * sigma c >= 0

显然只需要改变一下边权，跑一遍最长路即可

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define double long double
const int N = 200200;
const double eps = 1e-15;
int n, m;
int deg[N];
struct Edge {int nxt;int to;int w1;int w2;Edge() {}Edge(int x, int y, int w, int z): nxt(x), to(y), w1(w), w2(z) {}
} e[N << 2];
int h[N], cnt;
void add(int u, int v, int w1, int w2) {e[++cnt] = Edge(h[u], v, w1, w2);h[u] = cnt;return;
}
double L = 0, R = 2e17;
double mx[N];
bool check(double x) {for (int i = 1; i <= n; ++i) {mx[i] = -2e17;}mx[1] = 0;for (int u = 1; u <= n; ++u) {for (int i = h[u]; i; i = e[i].nxt) {int v = e[i].to;double w1 = e[i].w1, w2 = e[i].w2;mx[v] = max(mx[v], mx[u] + w1 - x * w2);}}if (mx[n] >= 0)return 1;return 0;
}
signed main() {scanf("%d%d", &n, &m);for (int i = 1; i <= m; ++i) {int u, v, w1, w2;scanf("%d%d%d%d", &u, &v, &w1, &w2);add(u, v, w1, w2);deg[v]++;}while (R - L > eps) {double mid = (L + R) / 2.0;if (check(mid)) {L = mid;} else {R = mid;}}printf("%.18Lf\n", L);return 0;
}