PAT Advanced 1033 To Fill or Not to Fill (25 分)

PAT Advanced 1033 To Fill or Not to Fill (25 分)
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大家好,我是架构君,一个会写代码吟诗的架构师。今天说一说PAT Advanced 1033 To Fill or Not to Fill (25 分),希望能够帮助大家进步!!!

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


#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
    int a,b,sum=0,tmp;
    vector<int> vec1,vec2;
    cin>>a;
    while(a--){
        scanf("%d",&tmp);
        vec1.push_back(tmp);
    }
    cin>>b;
    while(b--){
        scanf("%d",&tmp);
        vec2.push_back(tmp);
    }
    sort(vec1.begin(),vec1.end(),greater<int>());
    sort(vec2.begin(),vec2.end(),greater<int>());
    int i=0;int j=0;
    while(vec1[i]>0&&vec2[j]>0){
        sum+=(vec1[i]*vec2[j]);
        i++;
        j++;
        if(i>vec1.size()-1||j>vec2.size()-1) break;
        //尽量少用erase,擦除线性表是需要一个复杂度的
    }
    i=vec1.size()-1;j=vec2.size()-1;
    while(vec1[i]<0&&vec2[j]<0){
        sum+=(vec1[i]*vec2[j]);
        i--;
        j--;
        if(i<0||j<0) break;
    }
    cout<<sum;
    system("pause");
    return 0;
}

 

转载于:https://www.cnblogs.com/littlepage/p/11355723.html

本文来源weixin_30595035,由架构君转载发布,观点不代表Java架构师必看的立场,转载请标明来源出处:https://javajgs.com/archives/29466

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