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 8 6.00 1250 7.00 600 7.00 150 7.10 0 7.20 200 7.50 400 7.30 1000 6.85 300 Sample Output 1: 749.17 Sample Input 2: 50 1300 12 2 7.10 0 7.00 600 Sample Output 2: The maximum travel distance = 1200.00 注：該貪心算法思想：設滿箱油時，車輛最遠行駛距離為maxd 按距離排序后，最初處于起點加油站記為now，在距離該站maxd范圍內，找出第一個比該站油價更低的站k，到k站加油，若找不到比now站低的站，就找距離該站maxd范圍內now站除外油價最低的站k，到該站加油，更新now為k
#include<cstdio>#include<algorithm>using namespace std;const int maxn=510;const int INF=1000000;struct station{ double price,dis;}st[maxn];bool cmp(station a,station b){ return a.dis<b.dis;} int main(){ int n; double Cmax,D,Davg; scanf("%lf%lf%lf%d",&Cmax,&D,&Davg,&n); for(int i=0;i<n;i++){ scanf("%lf%lf",&st[i].price,&st[i].dis); } st[n].price=0; st[n].dis=D; sort(st,st+n,cmp); if(st[0].dis!=0){ printf("The maximum travel distance = 0.00/n"); }else{ int now=0;//當前所處加油站編號 double ans=0,nowTank=0,MAX=Cmax*Davg;//ans是到達當前加油站時的最低花費，nowTank是到達當前加油站時油箱的油量 while(now<n){//每次循環將選出下一個需要到達的加油站 int k=-1;//選出的加油站編號 double priceMin=INF; for(int i=now+1;i<=n&&st[i].dis-st[now].dis<=MAX;i++){//選出從當前加油站滿油能到達范圍內的第一個油價低于當前//油價的加油站，如果沒有低于當前油價的加油站，則選擇價格最低的那個 if(st[i].price<priceMin){ priceMin=st[i].price; k=i; if(priceMin<st[now].price){ break; } } } if(k==-1) break; double need=(st[k].dis-st[now].dis)/Davg; if(priceMin<st[now].price){ if(nowTank<need){ ans+=(need-nowTank)*st[now].price; nowTank=0;//到達下一個加油站，更新油箱中的油量 }else{ nowTank-=need;//到達下一個加油站，更新油箱中的油量 } }else{ ans+=(Cmax-nowTank)*st[now].price; nowTank=Cmax-need;//到達下一個加油站，更新油箱中的油量 } now=k;//到下一個加油站 } if(now==n){//能夠到達終點站 printf("%.2f/n",ans); }else{ printf("The maximum travel distance = %.2f/n",st[now].dis+MAX); } } return 0;}新聞熱點
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