【BZOJ2876】[Noi2012]骑行川藏【二分】【拉格朗日乘数法】

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【题目链接】

显然是一个约束条件下多元函数最值问题，那么就用拉格朗日乘数法就行了。然后二分lambda的值，解方程。

具体见POPOQQQ大爷的题解

【POPOQQQ的题解】

/* Telekinetic Forest Guard */#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long double LD;const int maxn = 10005;const LD eps = 1e-12;int n;LD E, s[maxn], k[maxn], v[maxn], u[maxn];inline bool check(LD lambda) {LD res = 0.0;for(int i = 1; i <= n; i++) {LD l = max((LD)0.0, v[i]), r = 0x3f3f3f3f;while(r - l > eps) {LD mid = (l + r) / 2;if(2 * lambda * k[i] * mid * mid * (mid - v[i]) > 1.0) r = mid;else l = mid;}u[i] = r;res += k[i] * (u[i] - v[i]) * (u[i] - v[i]) * s[i];}return res > E;}int main() {scanf("%d%Lf", &n, &E);for(int i = 1; i <= n; i++) scanf("%Lf%Lf%Lf", &s[i], &k[i], &v[i]);LD l = 0.0, r = 0x3f3f3f3f;while(r - l > eps) {LD mid = (l + r) / 2;if(check(mid)) l = mid;else r = mid;}LD ans = 0.0;for(int i = 1; i <= n; i++) ans += s[i] / u[i];printf("%.10Lf\n", ans);return 0;}

~~庆幸看过数分~~

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