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 (≤), the distance between this station and Hangzhou, for ,. 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
1 #include<iostream> 2 #include<vector> 3 #include<algorithm> 4 const double inf = 99999999.0; 5 struct Node{ 6 double pri; 7 int dis; 8 }; 9 bool cmp(Node& a, Node& b){return a.dis<b.dis;} 10 using namespace std; 11 /* 12 *******注意输入double类型要用%lf 13 当确信自己的算法没有问题的时候， 还是有测试点出错 那可能是自己漏掉了信息 或者没考虑一些边界条件 14 */ 15 int main(){ 16 double cMax, dAvg; 17 int n, i, j, d; 18 scanf("%lf%d%lf%d", &cMax, &d, &dAvg, &n); 19 vector<Node> v(n); 20 for(i=0; i<n; i++)scanf("%lf%d", &v[i].pri, &v[i].dis); 21 Node node; 22 node.dis=d; node.pri=inf; 23 v.push_back(node); 24 sort(v.begin(), v.end(), cmp); 25 double sum=0, tank=0; 26 bool flag = true; 27 //坑！ 出发点可能没有油站 不能出发 28 if(v[0].dis!=0){ printf("The maximum travel distance = 0.00"); return 0;} 29 for(i=0; i<n; ){ 30 if(v[i+1].dis-v[i].dis>cMax*dAvg){ 31 flag = false; 32 printf("The maximum travel distance = %.2f", v[i].dis+cMax*dAvg); 33 break; 34 }else{ 35 if(v[i+1].pri<=v[i].pri){ 36 sum += ((v[i+1].dis-v[i].dis)/dAvg-tank)*v[i].pri; 37 tank=0; 38 i++; 39 }else{ 40 int minidx=i+1; //找后面续航范围内最便宜的油站 41 for(j=i+1; j<n&&((v[j].dis-v[i].dis)<=cMax*dAvg); j++) 42 if(v[minidx].pri>=v[j].pri) minidx = j; 43 if(v[i].pri<=v[minidx].pri){ 44 if((d-v[i].dis)<=cMax*dAvg){//该站到终点距离小于最大续航能力 45 sum += ((d - v[i].dis)/dAvg-tank)*v[i].pri; 46 break; 47 }else{//该站到终点距离大于续航能力 把油加满， 到续航范围内最便宜的加油站加油 48 sum += (cMax - tank)*v[i].pri; 49 tank = cMax - (v[minidx].dis - v[i].dis)/dAvg; 50 } 51 }else{ 52 sum += ((v[minidx].dis - v[i].dis)/dAvg-tank)*v[i].pri; 53 tank = 0; 54 } 55 i = minidx; 56 } 57 } 58 } 59 if(flag) printf("%.2f", sum); 60 return 0; 61 }