三分 --- CSU 1548: Design road

 Design road

Problem's Link:   http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1548

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Mean: 

目的：从(0,0)到达(x,y)。但是在0~x之间有n条平行于y轴的河，每条河位于xi处，无限长，wi宽，并分别给出了建立路和桥每公里的单价

求：到达目标的最小费用。

analyse:

比赛的时候一直没想到思路，第二个样列怎么算都算不对，赛后才知道是三分。

首先把所有的桥移到最右端，然后三分枚举路和河的交点。

Time complexity: O(logn)

Source code: 

//  Memory   Time
//  1347K     0MS
//   by : crazyacking
//   2015-03-30-21.24
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<iostream>
#include<algorithm>
#define MAXN 1000010
#define LL long long
using namespace std;
double x,y,c1,c2,sum,xx;
double calc(double mid)
{
        double road_cost=sqrt(xx*xx+mid*mid)*c1, bridge_cost=sqrt(sum*sum+(y-mid)*(y-mid))*c2;
        return road_cost+bridge_cost;
}

double solve(double low,double high)
{
        double l=low,h=high;
        double mid=(l+h)/2,mmid=(mid+h)/2;
        double cmid=calc(mid),cmmid=calc(mmid);
        while(fabs(cmmid-cmid)>=1e-10)
        {
                if(cmid>cmmid)
                        l=mid;
                else
                        h=mmid;
                mid=(l+h)/2,mmid=(mid+h)/2;
                cmid=calc(mid),cmmid=calc(mmid);
        }
        return min(cmmid,cmid);
}

int main()
{
        int n;
        while(cin>>n>>x>>y>>c1>>c2)
        {
                sum=0.0;
                double tmp1,tmp2;
                for(int i=1;i<=n;++i)
                {
                        cin>>tmp1>>tmp2;
                        sum+=tmp2;
                }
                xx=x-sum;
                printf("%.2lf
",solve(0.0,y));
        }
        return 0;
}