CSU 1548 Design road（三分查找）

Posted 2020-09-12 Dh_q

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。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <vector>
 7 #include <map>
 8 #include <set>
 9 #include <queue>
10 #include <sstream>
11 #include <cmath>
12 #include <algorithm>
13 using namespace std;
14 #define pb push_back
15 #define mp make_pair
16 #define exp 1e-7
17 #define ms(a, b)  memset((a), (b), sizeof(a))
18 //#define LOCAL
19 typedef long long LL;
20 const int inf = 0x3f3f3f3f;
21 const int maxn = 200000+10;
22 const int mod = 1000000007;
23 double n, x, y, c1, c2, riverlen, landlen;
24 double cut(double hh)
25 {
26     return c2*hypot(riverlen, hh) + c1*hypot(landlen, y-hh);
27 }
28 int main() {
29 #ifdef LOCAL
30     freopen("input.txt" , "r", stdin);
31 #endif // LOCAL
32     while(~scanf("%lf%lf%lf%lf%lf", &n, &x, &y, &c1, &c2)){
33         riverlen = 0.0;
34         for(int i=0;i<n;i++){
35             double xi, wi;
36             scanf("%lf%lf", &xi, &wi);
37             riverlen += wi;
38         }
39         landlen = x - riverlen;
40         double l , r, mid, mmid;
41         l = 0;
42         r = y;
43         while( l + exp < r){
44             mid = ( l + r ) / 2;
45             mmid = (mid + r) / 2;
46             if( cut(mid) <= cut(mmid) )
47                 r = mmid;
48             else
49                 l = mid;
50         }
51         printf("%.2f\n", cut(l));
52     }
53     return 0;
54 }

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