（三分）Design road，湖南多校对抗赛-优快云博客

本文介绍了一种算法，用于在存在多条平行河流的情况下，计算从起点到终点建造道路及桥梁所需的最低成本。通过三分法搜索最优高度，实现对河流与陆地交界处的最佳路径规划。

Design road

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 9, Accepted users: 9

Problem 13267 : Special judge

Problem description

You need to design road from (0, 0) to (x, y) in plane with the lowest cost. Unfortunately, there are N Rivers between (0, 0) and (x, y).It costs c1 Yuan RMB per meter to build road, and it costs c2 Yuan RMB per meter to build a bridge. All rivers are parallel to the Y axis with infinite length.

Input

There are several test cases.

Each test case contains 5 positive integers N,x,y,c₁,c₂ in the first line.(N ≤ 1000,1 ≤ x,y≤ 100,000,1 ≤ c₁,c₂ ≤ 1000).

The following N lines, each line contains 2 positive integer x_i, w_i ( 1 ≤ i ≤ N ,1 ≤ x_i < x, x_i-1+w_i-1 < x_i , x_N+w_N ≤ x),indicate the i-th river(left bank) locate x_i with w_i width.

The input will finish with the end of file.

Output

For each the case, your program will output the least cost P on separate line, the P will be to two decimal places .

Sample Input

1 300 400 100 100
100 50
1 150 90 250 520
30 120

Sample Output

50000.00
80100.00

Judge Tips

——Dr.Wu

Problem Source

HNU Contest

分析：合并一下所有河流与所有陆地，三分河流与陆地交界线的高度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 1005;
typedef long long LL;
const int INF = 0xfffffff;
const double EPS = 1e-8;
double N, x, y, c1, c2;
double River;
double f(double h)
{
    return sqrt(River*River + h*h)*c2 + sqrt((x - River)*(x - River) + (y-h)*(y-h))*c1;
}
double ternary(double L, double R)
{
    while (R - L > EPS)
    {
        double mid1 = L + (R - L) / 3;
        double mid2 = R - (R - L) / 3;
        if (f(mid1) < f(mid2))
            R = mid2;
        else
            L = mid1;
    }
    return f((L + R) / 2);
}
int main()
{
    //freopen("f:\\input.txt", "r", stdin);
    while (~scanf("%lf%lf%lf%lf%lf", &N, &x, &y, &c1, &c2))
    {
        River = 0;
        for (int i = 0; i < N; i++)
        {
            double l, w;
            scanf("%lf%lf", &l, &w);
            River += w;
        }
        double ans = ternary(0, y);
        printf("%.2f\n", ans);
    }
    return 0;
}