CSU 1548 Road And Bridge acm 三分

最新推荐文章于 2017-10-21 20:33:49 发布

qr_blog

最新推荐文章于 2017-10-21 20:33:49 发布

阅读量359

点赞数

CC 4.0 BY-SA版权

分类专栏：三分文章标签：三分

本文链接：https://blog.youkuaiyun.com/qr_blog/article/details/47273687

三分专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种解决从起点到终点经过多个平行于Y轴的河流的路径规划问题的方法。通过合并所有河流并利用三分法找到从陆地到河流最优路径的最低成本方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

首先将所有的河流都合并在一边，然后在陆地与边上任找一点，计算相应的花费，用三分的方法得到最小的花费

#include<iostream>
#include<iomanip>
#include<math.h>
using namespace std;
double n, x, y, c1, c2, p[1010][2], sum, m;
double dis(double x1, double y1,double x2,double y2){
double mm = sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
return mm;
}
double cal(double mm){
double nn = c2*dis(0, 0, sum, mm)+c1*dis(sum,mm,x,y);
return nn;
}
int main(){
while (cin >> n >> x >> y >> c1 >> c2){
  for (int i = 0; i < n; i++){
   cin >> p[i][0] >> p[i][1];
  }
  sum = 0;
  for (int i = 0; i < n; i++){
   sum += p[i][1];
  }
  double low = 0, high = y, mid = (low + high) / 2, midmid;
  while (low + 1e-6 < high){
   mid = (low + high) / 2;
   midmid = (mid + high) / 2;
   if (cal(mid) < cal(midmid))high = midmid;
   else low = mid;
  }
  cout << fixed << setprecision(2) << cal(mid) << endl;
}
return 0;
}