簡單易懂 PAT 1033 To Fill or Not to Fill 貪心算法

取消註釋， 看跑路過程！

 

貪心思路：

1. 優先前往油價比當前油價更低的油站，爲了前往這個油站，油箱裏的油要加最少（即跑到下一站油量恰好爲0），爲了能去那裏加更便宜的油

2. 沒有比當前油價低的油站時，則在能跑到的油站裏尋找價格最低的油站，但是這個最低的油站也比當前油站的價格高，所以要在當前油站把油加滿。

3. 如果滿油都到不了任意一個油站，那就說明到不了目的地，最遠到達當前油站座標+滿油箱能跑的距離

 

#include <iostream>
#include <algorithm>

using namespace std;
struct Station{
    double price;
    double dis;
};
bool cmp(Station a, Station b){
    return a.dis < b.dis;
}

int main(){
    double cap;
    double dis;
    double per;
    int n;
    scanf("%lf %lf %lf %d", &cap, &dis, &per, &n);
    struct Station sta[n+1];
    for(int i=0; i<n; ++i){
        scanf("%lf %lf", &sta[i].price, &sta[i].dis);
    }
    sta[n].price = 0;
    sta[n].dis = dis;
    n++;
    sort(sta, sta+n, cmp);
    //printf("\n");for(int i=0; i<n; ++i){printf("%d %.2f %.2f\n",i,sta[i].dis, sta[i].price);}printf("\n");

    if(sta[0].dis!=0){
        printf("The maximum travel distance = 0.00\n");
        return 0;
    }
    int now_sta = 0;
    int next_sta = 0;
    double V=0;
    double money=0;
    while(now_sta<n-1){
        for(int i=now_sta; i!=n; ++i){
            if(sta[i].dis-sta[now_sta].dis<=per*cap){
                if(sta[i].price<sta[now_sta].price){
                    next_sta = i;
                    if( (sta[i].dis - sta[now_sta].dis)/per >= V){
                        //printf("當前車站%d 油箱剩餘油量%.3f 前往車站%d 油不夠，加油%.3f\n",now_sta,V,next_sta,(sta[i].dis - sta[now_sta].dis)/per - V);
                        money += ((sta[i].dis - sta[now_sta].dis)/per - V) * sta[now_sta].price;
                        V=0;
                        break;
                    }else{
                        //printf("當前車站%d 油箱剩餘油量%.3f 前往車站%d 油夠，要耗油%.3f\n",now_sta,V,next_sta,(sta[i].dis - sta[now_sta].dis)/per - V);
                        V -= (sta[i].dis - sta[now_sta].dis)/per;
                        break;
                    }
                }
            }else{
                break;
            }
        }
        if(next_sta == now_sta){
            double min_price = std::numeric_limits<double>::max();
            for(int i=now_sta+1; i<n; ++i){
                if(sta[i].dis-sta[now_sta].dis<=per*cap){
                    if(sta[i].price < min_price){
                        next_sta = i;
                        min_price = sta[i].price;
                    }
                }else{
                    break;
                }
            }
            if(next_sta == now_sta){
                printf("The maximum travel distance = %.2f\n", sta[now_sta].dis+per*cap);
                return 0;
            }else{
                //printf("當前車站%d 油箱剩餘油量%.3f 前往車站%d 要加滿，加了：%.3f\n",now_sta,V,next_sta,50-(sta[next_sta].dis - sta[now_sta].dis)/per);
                money += (cap - V) * sta[now_sta].price;
                V = 50 - (sta[next_sta].dis - sta[now_sta].dis)/per;
            }
        }
        //printf("%d %d\n",now_sta,next_sta);
        now_sta = next_sta;
    }
    printf("%.2f\n",money);
    return 0;
}