题目1437：To Fill or Not to Fill

题目1437：To Fill or Not to Fill

时间限制：1 秒

内存限制：128 兆

(贪心有点复杂)

题目描述：

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.

输入：

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 (<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.

输出：

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.

样例输入：

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
50 1300 12 2
7.10 0
7.00 600

样例输出：

749.17
The maximum travel distance = 1200.00

#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;

class Station
{
public:
    int dis;
    float pri;
    Station()
    {
        dis = 0;
        pri = 0;
    }

};

bool cmp(Station a, Station b)
{
    return a.dis < b.dis;
}

Station s[501];
float res;//结果


int main()
{

    int cmax, d,  n, i, j;
    float davg;
    float maxDis;//加满一次行驶最大距离
    float res = 0;

    while(cin >> cmax)
    {
        res = 0;
        cin >> d >> davg >> n;
        maxDis = cmax * davg;
        for (i=0; i<n; i++)
        {
            cin >> s[i].pri >> s[i].dis;
        }
        s[n].dis = d;//去特殊化
        sort(s, s+n, cmp);
        if (s[0].dis != 0)
        {
            cout << "The maximum travel distance = 0.00" << endl;
            continue;
        }

        int curPos=0;
        float curOil=0;//当前距离、油箱剩油
        for (i=0; i<n ; i++)//每次一站
        {
            curPos = s[i].dis;
            //无法到达终点
            if (curPos + maxDis < s[i+1].dis)
            {
                cout << "The maximum travel distance = ";
                cout << setiosflags(ios::fixed) << setprecision(2) << curPos+maxDis +0.0<< endl;
                break;
            }
            //减掉中间行驶耗油
            if (i>0)
            {
                curOil = curOil - (s[i].dis - s[i-1].dis)/davg;
            }
            //如果最大距离内有较便宜
            int curDis = 0;
            bool isCheap = false;
            for (j=i+1; j<n; j++)
            {
                curDis = s[j].dis - s[i].dis;
                if (curDis>maxDis)//如果没有
                {
                    //res += (cmax - curOil)*s[i].pri;
                    //curOil = cmax;
                    break;
                }
                if (s[j].pri<s[i].pri)//有行驶到便宜站
                {
                    float temp = curOil;

                    curOil = (s[j].dis - s[i].dis )/davg;//应保证油量
                    if (temp - curOil < 0)
                    {
                        res += (curOil - temp)*s[i].pri;//补充加油
                    }
                    else
                    {
                        curOil = temp;//不需要加油
                    }
                    isCheap = true;
                    break;
                }
            }// end of for
            //最大距离内未找到便宜站
            if (!isCheap)
            {
                if (curPos+maxDis>=d)//到达终点
                {
                    res += ((d - curPos)/davg - curOil) *s[i].pri;
                    cout << setiosflags(ios::fixed) << setprecision(2);
                    cout << res << endl;

                    break;
                }
                else//未到终点，加满油
                {
                    res += (cmax - curOil)*s[i].pri;
                    curOil = cmax;
                }
            }
        }//end of for （每次一站）

    }//end of while
    return 0;
}