Design road

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

---

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)/,mmid=(mid+h)/;

        double cmid=calc(mid),cmmid=calc(mmid);

        while(fabs(cmmid-cmid)>=1e-)

        {

                if(cmid>cmmid)

                        l=mid;

                else

                        h=mmid;

                mid=(l+h)/,mmid=(mid+h)/;

                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=;i<=n;++i)

                {

                        cin>>tmp1>>tmp2;

                        sum+=tmp2;

                }

                xx=x-sum;

                printf("%.2lf\n",solve(0.0,y));

        }

        return ;

}

三分 --- CSU 1548: Design road的相关教程结束。