1033. To Fill or Not to Fill (25)

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains 4 positive numbers: C_max(<= 100), the maximum capacity of the tank; D (<=30000), the distance between Hangzhou and the destination city; D_avg (<=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: P_i, the unit gas price, and D_i (<=D), the distance between this station and Hangzhou, for i=1,...N. All the numbers in a line are separated by a space.

Output Specification:

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.

50 1300 12 86.00 12507.00 6007.00 1507.10 07.20 2007.50 4007.30 10006.85 300Sample Output 1:
749.17Sample Input 2:
50 1300 12 27.10 07.00 600Sample Output 2:
The maximum travel distance = 1200.00
此题典型的贪心算法，可用直接解法，也可用深度遍历。
#include<cstdio>#include<algorithm>using namespace std;const double INF = 100000000;struct Station{	double gas_price;	double dis;}st[505];bool cmp(Station a, Station b){	return a.dis < b.dis;}int main(){	double Cmax, Dis_total, Dis_Avg;	int n;	scanf("%lf %lf %lf %d", &Cmax, &Dis_total, &Dis_Avg, &n); 	for(int i = 0; i < n; ++i)	{		scanf("%lf %lf", &st[i].gas_price, &st[i].dis);	}	st[n].gas_price = 0;	st[n].dis = Dis_total;	sort(st, st + n, cmp);	if(st[0].dis != 0)	{		printf("The maximum travel distance = 0.00/n");	}else{		int now = 0;//当前加油站编号		//总油钱, 装满油能跑最远距离 , 当前油量		double ans = 0, MAX = Dis_Avg * Cmax, now_gas = 0;		while(now < n)		{			int k = -1;			double price_min = INF;			for(int i = now + 1; i <= n && st[i].dis <= st[now].dis + MAX; ++i)//在加满油能到达的加油站内遍历，找油价尽量低的加油站			{				if(st[i].gas_price < price_min)				{					price_min = st[i].gas_price;					k = i;					if(price_min < st[now].gas_price) break;				}			} 			if(k == -1) break;  //油满状态下到不了加油站，退出循环			//当能到达下一加油站，计算转移费用			double need = (st[k].dis - st[now].dis) / Dis_Avg;//从now站到k站需要的油			if(price_min < st[now].gas_price)//如果要k站油价比now站低，加到刚好能到达k站的油			{				if(need > now_gas)//如果当前油不够去k站的				{					ans += (need - now_gas) * st[now].gas_price;//加满刚好到k站的油					now_gas = 0;//到k站后油为0				}else{  //油量够去k站，就直达k站					now_gas -= need;				}			}else{  //如果k站油和now站油价相同或比now站高 加满油				ans += (Cmax - now_gas) * st[now].gas_price;  //加满到k站油				now_gas = Cmax - need; //剩下油为Cmax减去从now站到k站消耗的			}			now = k; //当请已到k站		} 		if(now == n)//可以到终点站			printf("%.2f/n", ans);		else 			printf("The maximum travel distance = %.2f/n", st[now].dis + MAX);//从now站加满油也到不了下一站  输出最远到达的距离	}	return 0;}