题目1437：To Fill or Not to Fill：贪心算法解决加油站选择问题（未解决）

//贪心算法解决加油站选择问题
//# include<iostream>
# include<stdio.h>
using namespace std;

# include<algorithm>

struct Node
{
    float p, d;
};
bool cmp(Node a, Node b)
{
    return a.d < b.d;
}

int main()
{
    Node node[501];
    float Cmax, D, Davg, distance, price, Ccur, Pcur, j;//double 会出问题，蛋疼
    int N, Ncur, i, k, flag;
    //while (cin >> Cmax >> D >> Davg >> N)
    while (scanf_s("%f%f%f%d", &Cmax, &D, &Davg, &N) != EOF)
    {
        for (i = 1; i <= N; i++)
        {
            //cin >> node[i].p >> node[i].d;
            scanf_s("%f%f", &node[i].p, &node[i].d);
        }
        //sort station by distance  
        sort(node + 1, node + 1 + N, cmp);
        
        if (N == 0 || node[1].d > 0.00001 || node[1].d < -0.00001)//第一个站点不在距离0处
        {
            printf("The maximum travel distance = 0.00
");
        }
        else//greedy now
        {
            price = 0;//当前花费
            distance = 0;//当前距离
            Ncur = 1;//当前所在加油站
            Ccur = 0;//当前汽油数量
            Pcur = node[1].p;//当前加油站的油价
            while (Ncur <= N)
            {
                //1.using 【remain】 gas find 【nearest】 【cheaper】 station
                flag = -1;
                for (i = Ncur + 1; i <= N&&node[i].d <= distance + Ccur*Davg; i++)
                {
                    if (node[i].p < Pcur)
                    {
                        flag = i;
                        break;
                    }
                }
                if (flag != -1)//find a station,and get there
                {
                    //price 未变
                    Ccur -= (node[flag].d - distance) / Davg;//当前汽油数量===注意与下面表达式的计算顺序
                    distance = node[flag].d;//当前距离
                    Ncur = flag;//当前所在加油站
                    Pcur = node[flag].p;//当前加油站的油价
                    continue;
                }
                //else 1: 2.using 【Cmax】 find 【nearest】 【cheaper】 station
                flag = -1;
                for (i = Ncur + 1; i <= N&&node[i].d <= distance + Cmax*Davg; i++)
                {
                    if (node[i].p < Pcur)
                    {
                        flag = i;
                        break;
                    }
                }
                if (flag != -1)//find a station,and get there
                {
                    price += ((node[flag].d - distance) / Davg - Ccur)*node[Ncur].p;//当前花费
                    distance = node[flag].d;//当前距离
                    Ncur = flag;//当前所在加油站
                    Pcur = node[flag].p;//当前加油站的油价
                    Ccur = 0;//当前汽油数量：到K后的汽油为0
                    continue;
                }
                //else 2: 3.get the Cmax and go as far as possible
                flag = -1;
                for (i = Ncur + 1; i <= N&&node[i].d <= distance + Cmax*Davg; i++)
                {
                    flag = i;
                }
                if (flag != -1)//can get some one station,and get there
                {
                    price += (Cmax - Ccur)*node[Ncur].p;//当前花费
                    distance = node[flag].d;//当前距离
                    Ncur = flag;//当前所在加油站
                    Pcur = node[flag].p;//当前加油站的油价
                    Ccur = Cmax - (node[flag].d - distance) / Davg;//当前汽油数量
                    continue;
                }
                else//can not get some one station,over
                {
                    printf("The maximum travel distance = %.2f
", distance + Cmax*Davg);
                    break;
                }
            }
            if (Ncur > N)
            {
                printf("%.2lf
", price);
            }

        }
    }
    return 0;
}