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
此題典型的貪心算法，可用直接解法，也可用深度遍歷。
#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;//當(dāng)前加油站編號(hào)		//總油錢, 裝滿油能跑最遠(yuǎn)距離 , 當(dāng)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)//在加滿油能到達(dá)的加油站內(nèi)遍歷，找油價(jià)盡量低的加油站			{				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;  //油滿狀態(tài)下到不了加油站，退出循環(huán)			//當(dāng)能到達(dá)下一加油站，計(jì)算轉(zhuǎn)移費(fèi)用			double need = (st[k].dis - st[now].dis) / Dis_Avg;//從now站到k站需要的油			if(price_min < st[now].gas_price)//如果要k站油價(jià)比now站低，加到剛好能到達(dá)k站的油			{				if(need > now_gas)//如果當(dāng)前油不夠去k站的				{					ans += (need - now_gas) * st[now].gas_price;//加滿剛好到k站的油					now_gas = 0;//到k站后油為0				}else{  //油量夠去k站，就直達(dá)k站					now_gas -= need;				}			}else{  //如果k站油和now站油價(jià)相同或比now站高 加滿油				ans += (Cmax - now_gas) * st[now].gas_price;  //加滿到k站油				now_gas = Cmax - need; //剩下油為Cmax減去從now站到k站消耗的			}			now = k; //當(dāng)請(qǐng)已到k站		} 		if(now == n)//可以到終點(diǎn)站			printf("%.2f/n", ans);		else 			printf("The maximum travel distance = %.2f/n", st[now].dis + MAX);//從now站加滿油也到不了下一站  輸出最遠(yuǎn)到達(dá)的距離	}	return 0;}