好像是 01 分數規劃 板子

列出式子 (F - ∑C_i) / (∑T_i)

設當前二分的答案為 ans

若 ans ≤ (F - ∑C_i) / (∑T_i)
則說明 ans 可以再增大

那我們二分邊界調整的條件就有了

就是當 F - (∑ans*T_i + ∑C_i) ≥ 0

代碼：

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

const int MAXN = 405, MAXM = 10005;

struct EDGE {
    int x, y, c, t;
    double hf;
    bool operator < (const EDGE& b) const {
        return hf < b.hf;
    }
}edge[MAXM];
int n, m, f;
int fa[MAXN];

inline void clearfa() {
    for (int i = 1; i <= n; ++i) fa[i] = i;
}
int findfa(int x) {
    return ((fa[x] == x) ? (x) : (fa[x] = findfa(fa[x])));
}
inline bool link(int x, int y) {
    register int fx = findfa(x), fy = findfa(y);
    if (fx == fy) return false;
    fa[fx] = fy;
    return true;
}
inline bool chk(double mid) {
    register double ans = 0.0;
    clearfa();
    for (int i = 1; i <= m; ++i) edge[i].hf = edge[i].t * mid + edge[i].c;
    sort(edge + 1, edge + m + 1);
    for (int i = 1; i <= m; ++i) {
        if (link(edge[i].x, edge[i].y)) {
            ans += edge[i].hf;
        }
    }
    return ((ans - (double)f) <= 0.0);
}
inline void hfs(double l, double r) {
    register double mid = 0.0;
    while ((r - l) > 1e-7) {
        mid = ((l + r) / 2.0);
        if (chk(mid)) l = mid;
        else r = mid;
    }
    printf("%.4lf\n", l);
}

int main() {
    scanf("%d%d%d", &n, &m, &f);
    for (int i = 1; i <= m; ++i) {
        scanf("%d%d%d%d", &edge[i].x, &edge[i].y, &edge[i].c, &edge[i].t);
    }
    hfs(0, 2000000000.0);
    return 0;
}

[USACO 2001 OPEN] 地震