题目描写叙述：

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.

例子输入：

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

例子输出：

749.17
The maximum travel distance = 1200.00

分析：

贪心策略：如果如今自己处于A站，要考虑的是A站要不要加油，加多少油的问题。找到当前可达范围内（距离A站cmax*davg）下一个要加油的站B。

A站可达范围内， 分三种情况：

①没有加油站，------- 快到终点了，则加适量油到终点。或者 Impossible。则A站加满油到哪算哪；

②有更廉价的加油站 ------- 则找到第一家比A廉价的加油站B，加尽可能少的油（也可能油够直接开过去）到B站。

③仅仅有价格更高的加油站，------则当下A站加满油，寻找相对最廉价的加油站B 开过去。

对每个加油站进行贪心选择。

代码：

#include
#include
#include
#include
#include

using namespace std;

struct gas_station{
float price;
int distance;

bool operator < (const gas\_station & A)const {  
    return distance < A.distance;  
}

}buff[501];
int main()
{
int Cmax,D,Davg,N;
int dis;
while(scanf("%d %d %d %d",&Cmax,&D,&Davg,&N)!=EOF){
for(int i = 0; i < N; i++){
scanf("%f %d",&buff[i].price,&buff[i].distance);
}
buff[N].distance = D;
buff[N].price = 100000000;

    sort(buff,buff+N);

    if(buff\[0\].distance > 0){  
        printf("The maximum travel distance = 0.00\\n");  
        continue;  
    }

    dis = Cmax \* Davg;//最大单次行车距离  
    float sum = 0;//总费用  
    float temp,remind\_gas = 0;//剩余汽油量  
    int i ,k;  
    for(i = 0; i < N;i++){  
        k = i+1;  
        if(i != 0){  
            remind\_gas -= (float)(buff\[i\].distance-buff\[i-1\].distance)/Davg;  
        }  
        //找到当前站之后第一个价格比它低的加油站  
        for(;k < N && buff\[i\].price <= buff\[k\].price;k++ )  
            continue;  
        if(buff\[k\].distance - buff\[i\].distance > dis){  
            sum += (Cmax-remind\_gas)\*buff\[i\].price;  
            remind\_gas = Cmax;  
        }else{  
            temp = (float)(buff\[k\].distance-buff\[i\].distance)/Davg - remind\_gas;  
            if(fabs(temp)>1e-5&&temp >0){  
                sum += temp\*buff\[i\].price;  
                remind\_gas = (float)(buff\[k\].distance-buff\[i\].distance)/Davg;  
            }  
        }

        if(buff\[i+1\].distance - buff\[i\].distance>dis){  
            double total\_d = (double)(buff\[i\].distance + dis);  
            printf("The maximum travel distance = %.2f\\n",total\_d);  
            break;  
        }  
    }  
    if(i==N){  
        printf ("%.2f\\n", sum);  
    }

}  
return 0;  

}

注意：

比較浮点数是否大于零，须要加上fabs（temp），由于这个地方还wa了一次。