CSU - 1548 三分找最值

Design road

Time Limit: 2000MS

Memory Limit: 262144KB

64bit IO Format: %lld & %llu

Submit Status

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

Source

题意：有河流的地方需要建桥每米花费c2，没有的需要建路每米花费c1,使得从（0,0）到（x,y)花费最少。

思路：合并所有河流记为sum，然后在[0,y]里面通过三分寻找一个值y1，使得从（0,0）到（sum，y1）加上（sum，y1）到（x，y）的花费最少。（把河流和道路分开易于寻找最值）

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
using namespace std;
#define eps 1e-3
double distance(double x1,double y1,double x2,double y2)//求两点之间的距离
{
return sqrt((double)(x1-x2)*(x1-x2) + (double)(y1-y2)*(y1-y2));
}
int main()
{
#ifdef CDZSC_June
freopen("t.txt","r",stdin);
#endif
int n;
double x,y,c1,c2,s,w,sum;
while(~scanf("%d%lf%lf%lf%lf",&n,&x,&y,&c1,&c2))
{
sum = 0;
for(int i =0; i<n; i++)
{
scanf("%lf%lf",&s,&w);
sum += w;//合并河流
}
sum = x - sum;//获取需要建路的宽度
double L = 0,R = y,mid,midmid;
while(R - L > eps)
{
mid = (L+R)/2;
midmid = (mid+R)/2;
double s1 = distance(0,0,sum,mid)*c1+distance(sum,mid,x,y)*c2;
double s2 =  distance(0,0,sum,midmid)*c1+distance(sum,midmid,x,y)*c2;
//midmid必定大于mid，所以当s1>s2时，意味着最小值在靠近s2的那一边，所以令L = mid来缩小最小值的寻找范围，反之同理
if(s1 > s2){
L = mid;
}
else{
R = midmid;
}
}
printf("%.2f\n",distance(0,0,sum,mid)*c1+distance(sum,mid,x,y)*c2);
}
return 0;
}