Pat(Advanced Level)Practice--1033(To Fill or Not to Fill)
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Pat1033代码
With highways available, driving a car from Hangzhou to any other city is easy. But since the tank capacity of a car is limited, we have to find gas stations on the way from time to time. Different gas station may give different price. You are asked to carefully design the cheapest route to go.
Input Specification:
Each input file contains one test case. For each case, the first line contains 4 positive numbers: Cmax (<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; Davg (<=20), the average distance per unit gas that the car can run; and N (<= 500), the total number of gas stations. Then N lines follow, each contains a pair of non-negative numbers: Pi, the unit gas price, and Di (<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the cheapest price in a line, accurate up to 2 decimal places. It is assumed that the tank is empty at the beginning. If it is impossible to reach the destination, print "The maximum travel distance = X" where X is the maximum possible distance the car can run, accurate up to 2 decimal places.
Sample Input 1:50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 300Sample Output 1:
749.17Sample Input 2:
50 1300 12 27.10 07.00 600Sample Output 2:
The maximum travel distance = 1200.00
#include<cstdio>#include<vector>#include<algorithm>using namespace std;typedef struct Gas{float price;float distance;}Gas;bool cmp(const Gas &l,const Gas &r){if(l.distance<r.distance)return true;elsereturn false;}int main(int argc,char *argv[]){float c,d,a;int n,i,j;float fee;vector<Gas> v;Gas temp;float maxdis;scanf("%f%f%f%d",&c,&d,&a,&n);for(i=0;i<n;i++){scanf("%f%f",&temp.price,&temp.distance);v.push_back(temp);}temp.price=0;temp.distance=d;v.push_back(temp);sort(v.begin(),v.end(),cmp);if(v[0].distance!=0){printf("The maximum travel distance = 0.00\n");return 0;}float left;maxdis=c*a;i=0;fee=0;while(v[i].distance<d){float minprice=-1;int mindex=-1,findex=-1;for(j=i+1;j<=n&&(v[j].distance-v[i].distance)<=maxdis;j++){if(minprice==-1||v[j].price<minprice){ //在maxdis范围内搜索price最小的加油站 minprice=v[j].price;mindex=j;}if(findex==-1&&v[j].price<v[i].price)//在maxdis范围内搜索第1个findex=j; //price小于目前价格的加油站}if(mindex==-1)//下一个加油站距离目前加油站的距离大于所能行驶的最大break; //距离maxdisif(findex==-1)//在maxdis范围内不存在小于目前价格的加油站{ //即最小价格的加油站也大于目前价格,此时应把油加满if(d-v[i].distance<=maxdis)//在maxdis距离内能到达终点{fee+=((d-v[i].distance)/a-left)*v[i].price;left=0;i=n;}else//不能到达终点,则加油使之能跑到maxdis范围内价格最小的加油站{fee+=(c-left)*v[i].price;left=c-(v[mindex].distance-v[i].distance)/a;i=mindex;}}else//maxdis范围内存在加油站价格小于当前价格,则应在本加油站加油使之{ //能跑到该加油站fee+=((v[findex].distance-v[i].distance)/a-left)*v[i].price;left=0;i=findex;}}if(v[i].distance==d)printf("%.2f\n",fee);elseprintf("The maximum travel distance = %.2f\n",v[i].distance+maxdis);return 0;}
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