CSU 1548 Design road（三分查找）

题目链接：https://cn.vjudge.net/problem/142542/origin

 

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 xi, wi ( 1 ≤ i ≤ N ,1 ≤ xi ≤x, xi-1+wi-1 < xi , xN+wN ≤ x),indicate the i-th river(left bank) locate xi with wi 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

Hint

 

题意：

　　　　给你一个二维的坐标系，你要从[0,0]走到[x,y]。但是其间会有一些平行于y轴的河，你需要架桥。给你每米的修路费，和每米的修桥费，问最少的费用是多少？

题解：

　　　　现场的时候看了这题，感觉是2元方程，很难求的感觉。赛后，师兄说是三分求最值。然后学了一下三分，就A了。

　　　　因为河全部是平行的，可以将河全部移动到一边，就可以变成一边是河，一边是陆地。

　　　　河的宽为：riverlen = 每一条和的宽度相加。陆地的宽为：landlen = x - riverlen。

　　　　设点[riverlen, yi]这个点时，费用最小。[0, 0]到[riverlen, yi]为桥的费用， 陆地的费用同理。

 

看了一下别人的代码，发现了一个hypot()函数;

查了一个是一个求直角三角形的斜边的函数　　double hypot(double x, double y)。新知识:D， 开心XD。

还有最主要就是三分的思想，到时会补一个三分详解。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <sstream>
#include <cmath>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define exp 1e-7
#define ms(a, b)  memset((a), (b), sizeof(a))
//#define LOCAL
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 200000+10;
const int mod = 1000000007;
double n, x, y, c1, c2, riverlen, landlen;
double cut(double hh)
{
    return c2*hypot(riverlen, hh) + c1*hypot(landlen, y-hh);
}
int main() {
#ifdef LOCAL
    freopen("input.txt" , "r", stdin);
#endif // LOCAL
    while(~scanf("%lf%lf%lf%lf%lf", &n, &x, &y, &c1, &c2)){
        riverlen = 0.0;
        for(int i=0;i<n;i++){
            double xi, wi;
            scanf("%lf%lf", &xi, &wi);
            riverlen += wi;
        }
        landlen = x - riverlen;
        double l , r, mid, mmid;
        l = 0;
        r = y;
        while( l + exp < r){
            mid = ( l + r ) / 2;
            mmid = (mid + r) / 2;
            if( cut(mid) <= cut(mmid) )
                r = mmid;
            else
                l = mid;
        }
        printf("%.2f\n", cut(l));
    }
    return 0;
}