[BZOJ2876][Noi2012]骑行川藏 && 二分+拉格朗日乘数

我智商不足 无法解释清楚这些东西

大爷题解传送门：http://blog.youkuaiyun.com/popoqqq/article/details/42366599

另一篇很不错的：http://www.cppblog.com/prime56/archive/2012/08/13/187087.aspx

大概就是lambda和vi是单调的然后二分一下

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
using namespace std;
typedef long long LL;
const int MAXN = 10000;
const double eps = 1e-12;
double x[MAXN+10], k[MAXN+10], s[MAXN+10], v[MAXN+10];
double E;
int n;
double check(double lambda) {
	double ret = 0;
	for(int i = 1; i <= n; i++) {
		double l = max(0.0, v[i]), r = 123456789;
		while(r-l > eps) {
			double mid = (l+r) / 2;
			if(2 * lambda * k[i] * mid * mid * (mid - v[i]) > 1) 
				r = mid;
			else l = mid;
		}
		x[i] = l;
		ret += s[i] * k[i] * (x[i]-v[i]) * (x[i]-v[i]);
	}
	return ret;
}
void solve() {
	double l = 0, r = 123456789;
	while(r-l > eps) {
		double mid = (l+r) / 2;
		if(check(mid) > E) 
			l = mid;
		else r = mid;
	}
}
int main() {
	SF("%d%lf", &n, &E);
	for(int i = 1; i <= n; i++) SF("%lf%lf%lf", &s[i], &k[i], &v[i]);
	solve();
	double ans = 0;
	for(int i = 1; i <= n; i++)
		ans += s[i] / x[i];
	PF("%.8f", ans);
}