1548: Design road

Submit Page Summary Time Limit: 2 Sec Memory Limit: 256 Mb Submitted: 450 Solved: 237

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.

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

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
using namespace std;
typedef long long LL;
#define max_v 1005
#define eps 1e-3
double dis(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
    int n;
    double x,y,c1,c2;
    while(~scanf("%d %lf %lf %lf %lf",&n,&x,&y,&c1,&c2))
    {
        double sum=0,w,x1;
        for(int i=0;i<n;i++)
        {
            scanf("%lf %lf",&x1,&w);
            sum+=w;
        }
        sum=x-sum;
        double L=0,R=y,mid,mmid;
        while(R-L>eps)
        {
            mid=(L+R)/2.0;
            mmid=(mid+R)/2.0;
            double s1=dis(0,0,sum,mid)*c1+dis(sum,mid,x,y)*c2;
            double s2=dis(0,0,sum,mmid)*c1+dis(sum,mmid,x,y)*c2;
            if(s1>s2)
                L=mid;
            else
                R=mmid;
        }
        printf("%.2lf\n",dis(0,0,sum,mid)*c1+dis(sum,mid,x,y)*c2);
    }
    return 0;
}
/*
題目意思：
給你兩點（0，0） （x，y）
在這兩點之間有n條平行於y軸的河流
修路每單位花費c1，搭橋每單位花費c2
問你到達（x，y）最小的花費是多少

分析：
把所有合河流移動到左邊，所以肯定是直接從（0，0）搭橋到河流對岸的某點（sum，y）
sum是所有河流的寬度
然後從（sum，y）修路到目的地
現在想象一下
整條路徑（包括路和橋）
路的兩端是固定的
中間河流對面某點（sum，y）是不固定的
隨著該點的移動，路徑長度的不同的
所以花費也是不同的
所以找一個合適的y
使得花費最小
極值尋找問題，采用三分
三分尋找合適的y
使得花費最小

註意精度問題
eps10的-3次方夠了
*/

1548: Design road （思維題 做法：三分找極值）