【题解】LuoGu4951：[USACO01OPEN]Earthquake

本文链接：https://blog.csdn.net/ModestCoder_/article/details/108424585

本文探讨了分数规划问题的解决方法，通过将问题转化为最小生成树问题，并使用二分法进行求解。介绍了如何根据给定条件调整边权，求得满足条件的最小生成树，从而得到分数规划问题的解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题传送门
01分数规划
$\frac{f-\sum c_x}{\sum t_x}=ans$
$f-\sum c_x-\sum t_x*ans=0$
$f>=\sum c_x+\sum t_x*ans$
二分答案 $m i d$ ，将边权赋为 $c_x+t_x*mid$
求最小生成树，总边权和 $f$ 比较

Code：

#include <bits/stdc++.h>
#define maxn 40010
using namespace std;
const double eps = 1e-6;
double F;
int f[maxn], n, m;
struct Line{
	int x, y;
	double c, t, l;
}line[maxn];

inline int read(){
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
} 

bool cmp(Line x, Line y){ return x.l < y.l; }
int getfa(int k){ return f[k] == k ? k : f[k] = getfa(f[k]); }

bool check(double mid){
	for (int i = 1; i <= n; ++i) f[i]  = i;
	for (int i = 1; i <= m; ++i) line[i].l = line[i].c + line[i].t * mid;
	sort(line + 1, line + 1 + m, cmp);
	double s = 0;
	for (int i = 1; i <= m; ++i){
		int s1 = getfa(line[i].x), s2 = getfa(line[i].y);
		if (s1 != s2) f[s1] = s2, s += line[i].l;
	}
	return F - s >= eps;
}

int main(){
	n = read(), m = read(), F = read();
	for (int i = 1; i <= m; ++i) line[i].x = read(), line[i].y = read(), line[i].c = read(), line[i].t = read();
	double l = 0, r = F, ans = 0;
	while (fabs(r - l) > eps){
		double mid = (l + r) / 2.0;
		if (check(mid)) ans = mid, l = mid; else r = mid;
	}
	printf("%.4lf\n", ans);
	return 0;
}