Can you find it? （三分搜索

最新推荐文章于 2021-11-06 16:23:23 发布

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最新推荐文章于 2021-11-06 16:23:23 发布

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分类专栏：二分&三分文章标签：三分

本文链接：https://blog.youkuaiyun.com/zzcblogs/article/details/70264335

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本文介绍了一个关于三数之和的问题及其算法实现，并探讨了一种涉及三分搜索的应用场景——寻找两点以达到最短路径的问题。文章通过具体实例展示了如何使用三分搜索来解决这类问题。

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Can you find it?

Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/10000 K (Java/Others)

Total Submission(s): 1199 Accepted Submission(s): 383

Problem Description

Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

Sample Input

Sample Output

Case 1:
NO
YES
NO

二分搜索搜索的是线性的，三分搜索搜索的是二次函数性质的（求最大值||最小值。

题意：

给你一个平面，平面上有两条线段AB和CD，某东西在AB上跑的速度是P，在CD上跑的速度是Q，在平面其他地方跑的速度是R，问从A点到D点的最快时间。

思路：

嵌套的三分搜索
1.首先明确需要在AB和CD上分别确定一个点作为转折点。
2.若已知AB，CD上的定点P1,P2,那么Time = AP1/P + P1P2/R + P2D/Q;
3.若我们能够先知道点P1(x1,y1),那么可以根据P2(x2,y2)求出最快的时间Time。
Time = AP1/P + sqrt((x1-x2)^2 + (y1-y2)^2)/R + sqrt((x2-D.x) + (y2-D.y))/Q;

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <stdio.h>
#include <vector>
#include <map>
#include <queue>
#include <utility>
typedef long long ll;
using namespace std;
struct Point {
    double x, y;
    Point(double x = 0.0, double y = 0.0):x(x), y(y){}
    Point& operator = (const Point& p) {
        this->x = p.x;
        this->y = p.y;
        return *this;
    }
};

Point A, B, C, D;

int P, Q, R;

Point Get_Point (const Point& x, const Point& y) {
    return Point((x.x + y.x)/2.0, (x.y + y.y)/2.0);
}

double Get_Len (const Point& x,const Point& y) {
    return sqrt((x.x - y.x)*(x.x - y.x) + (x.y - y.y)*(x.y - y.y));
}

double Get_Time (const Point& p) {
    Point left = C, right = D, mid, mmid;
    double mid_time = 0, mmid_time = 1;
    while (fabs(mmid_time - mid_time) > 1e-6) {
        mid = Get_Point(left, right);
        mmid = Get_Point(mid, right);
        
        mid_time = Get_Len(p, mid)/R + Get_Len(D, mid)/Q;
        mmid_time = Get_Len(p, mmid)/R + Get_Len(D, mmid)/Q;
        
        if (mid_time - mmid_time >= 1e-6)
            left = mid;
        else
            right = mmid;
    }
    return mid_time + Get_Len(A, p)/P;
}

int main () {
    int T;
    cin >> T;
    while (T--) {
        cin >> A.x >> A.y >> B.x >> B.y >> C.x >> C.y >> D.x >> D.y >> P >> Q >> R;
        
        Point left(A.x, A.y), right(B.x, B.y);
        double mid_time = 0, mmid_time = 1;
        
        while (fabs(mid_time - mmid_time) > 1e-6) {
            Point mid = Get_Point(left, right);
            Point mmid = Get_Point(mid, right);
            
            mid_time = Get_Time(mid);
            mmid_time = Get_Time(mmid);
            
            if (mid_time - mmid_time >= 1e-6)
                left = mid;
            else
                right = mmid;
        }
        printf("%.2lf\n", mid_time);
    }
}