csu 1548 Road And Bridge

Road And Bridge

思路：不要管河的位置都移到一边（右）+三分

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

//AC代码
#include<cstdio>
#include<cmath>
int  n;
double x1,y2,c1,c2,wid,real;
double fff(double y)
{
return c1*sqrt(real*real+y*y)+c2*sqrt(wid*wid+(y2-y)*(y2-y));
}
int main()
{
while(scanf("%d %lf %lf %lf %lf",&n,&x1,&y2,&c1,&c2)!=EOF)
{
int x,y;
double min,mid,midd,max;
wid=0;
while(n--)
{
scanf("%d%d",&x,&y);
wid+=y;
}
real=x1-wid;
min=0;
max=y2;
while((max-min)>1e-4)
{
mid=(min+max)/2;
midd=(mid+max)/2;
if(fff(mid)<fff(midd)) max=midd;
else min=mid;
}
printf("%.2lf\n",fff(mid));

}
return 0;
}