1033 To Fill or Not to Fill (25分)

Posted 2021-03-06 d-i-p

1. 题目

2. 思路

1. 把终点作为price=0， distance=d的点插入
2. 按照distance排序
3. 如果distance + c*davg内有station，找最近的比当前价格低的，买能恰好到达的gas，如果价格都高，买其中最低的，买满油
4. 如果distance + cdavg内没有station， 输出当前distance + cdavg

3. 注意点

1. 要设置当前油量的变量，可能不是0
2. 如果distance=0没有station， 输出0

4. tip

巧妙把最后d最为station插入， 减少了很多的讨论，使得代码变得简单，启发是看有些特殊情况能不能化为一般情况

5. 代码

#include<cstdio>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<iostream> 

// 21:51 - 23:12
using namespace std;

struct station{
    double price;
    double distance;
    station(double price, double distance){
        this->price = price;
        this->distance = distance;
    }
};

double c, d, davg;
int n;
vector<station> s;

bool cmp(station a, station b){
    return a.distance < b.distance;
}

int main(){
    scanf("%lf %lf %lf %d", &c, &d, &davg, &n);
    for(int i=0;i<n;i++){
        double price, distance;
        scanf("%lf %lf", &price, &distance);
        s.push_back(station(price, distance));
    }
    s.push_back(station(0, d));
    n++;
    sort(s.begin(), s.end(), cmp);
    if(s[0].distance != 0){
        printf("The maximum travel distance = 0.00");
        return 0;
    }
    
    double sum = 0;
    double now_c = 0;
    
    int i = 0;
    int pos = 0;
    double dis = c*davg; //一次最远距离 
    while(1){
        int index = -1;
        double min_price = 0x7fffffff;
        for(i=pos+1;i<n && s[i].distance-s[pos].distance<=dis;i++){
            if(s[i].price < s[pos].price){
                index = i;
                break;
            }
            if(s[i].price < min_price){
                min_price = s[i].price;
                index = i;
            }
        }
        if(i == n){
            printf("%.2f", sum);
            break;
        }
        if(index == -1){
            printf("The maximum travel distance = %.2f", s[pos].distance + dis);
            return 0;
        }
        if(s[pos].price < s[index].price){
            double need = (s[index].distance - s[pos].distance) / davg;
            sum += (c - now_c)*s[pos].price;
            now_c = c - need;
        }else{
            double need = (s[index].distance - s[pos].distance) / davg;
            if(need > now_c){
                sum += (need - now_c) * s[pos].price;
            }
            now_c = 0;
        }
        pos = index;
    }
}