三分 --- CSU 1548: Design road

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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 longusing 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\n",solve(0.0,y));        }        return 0;}

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