• 洛谷P2571 [SCOI2010]传送带


    题目链接:

    kma

    题目分析:

    裸的三分套三分啊,三分为什么是单峰的可以去看这篇博客的证明,感觉是目前写得最清楚的一篇

    →人赢FSYolanda吊打集训队

    不过其实最开始没有很搞懂三分求单峰函数最值是个啥东西,所以这里还是记录一下
    手画图,不要吐槽有多丑
    TIM图片20190808004033.jpg

    代码:

    #include <bits/stdc++.h>
    #define N (2000 + 10)
    using namespace std;
    inline int read() {
    	int cnt = 0, f = 1; char c = getchar();
    	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
    	return cnt * f;
    }
    int x, y;
    int P, Q, R;
    const double eps = 1e-8;
    int dcmp(double x) {
    	if (fabs(x) < eps) return 0;
    	else return x < 0 ? -1 : 1;
    }
    struct Vector{
    	double x; double y;
    	Vector (){};
    	Vector (double x_, double y_) : x (x_), y (y_) { };
    }A, B, C, D;
    typedef Vector Point;
    struct Line{
    	Point u; Point v;
    	Line (Point u_, Point v_) : u (u_), v(v_) { };
    };
    typedef Line Segment;
    
    Vector operator + (Vector A, Vector B) {
    	return Vector (A.x + B.x, A.y + B.y);
    }
    
    Vector operator - (Vector A, Vector B) { 
        return Vector (A.x - B.x, A.y - B.y);
    }
    
    Vector operator / (Vector A, double p) {
    	return Vector (A.x / p, A.y / p);
    }
    
    double Dot (Vector a, Vector b) {
    	return a.x * b.x + a.y * b.y;
    }
    double Length(Vector A){
        return sqrt(Dot(A, A));
    }
    
    double calc(Point X) {
    	Point l = C, r = D;
    	while (Length(l - r) > eps) {
    		Point x = (r - l) / 3;
    		Point lmid = l + x, rmid = r - x;
    		double ans1 = Length(lmid - D) / Q + Length(X - lmid) / R;
    		double ans2 = Length(rmid - D) / Q + Length(X - rmid) / R;
    		if (ans2 - ans1 > eps) r = rmid;
    		else l = lmid;
    	}
    	return Length(l - D) / Q + Length(X - l) / R;
    }
    double solve() {
    	Point l = A, r = B;
    	while (Length(l - r) > eps) {
    		Point x = (r - l) / 3;
    		Point lmid = l + x, rmid = r - x;
    		double ans1 = calc(lmid) + Length(lmid - A) / P;
    		double ans2 = calc(rmid) + Length(rmid - A) / P;
    		if (ans2 - ans1 > eps) r = rmid;
    		else l = lmid;
    	} return calc(l) + Length(l - A) / P;
    }
    int main() {
    	x = read(), y = read();
    	A = Point(x, y);
    	x = read(), y = read();
    	B = Point(x, y);
    	x = read(), y = read();
    	C = Point(x, y);
    	x = read(), y = read();
    	D = Point(x, y);
    	P = read(), Q = read(), R = read();
    	return printf("%0.2lf", solve()), 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/kma093/p/11318759.html
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