九度OJ 1437 To Fill or Not to Fill

题目大意：小明从杭州去往某目的地，要经过一些加油站，每个加油站的价格不一样。若能顺利到达，求加油费用最少为多少，否则求出能行驶的最远距离。

思路：贪心算法
1>若下一加油站的价格更便宜，则只需走到下一加油站即可。
2>若下一结点的价格没有该节点便宜
1.若将油箱加满，看看在其能到达的最远距离内，是否有比该点更便宜的站点。若有，则正好到达这个跟便宜的点即可；否则，将油箱加满，然后到达这段距离内价格最小的点（除当前点外）。

代码如下：

#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
struct station{
	double price;
	double dis;
}sta[502];

int cmp(const void *a,const void *b){
	station* p = (station *)a;
	station* q = (station *)b;
	return p->dis - q->dis;
}

int main(){
    double cmax,d,davg;
	int n,i,j;
	double nowgas,length,cost;
	while(scanf("%lf%lf%lf%d",&cmax,&d,&davg,&n) != EOF){
		nowgas = 0;
		length = 0;
		cost = 0;
		for(i=0; i<n; i++)
		    scanf("%lf%lf",&sta[i].price,&sta[i].dis);
		qsort(sta,n,sizeof(station),cmp);
		if(n == 0 || sta[0].dis != 0){
			printf("The maximum travel distance = 0.00
");
			continue;
		}
		sta[n].price = 0;
		sta[n].dis = d;
		for(i=0; i<n; i++){
			if(cmax*davg < sta[i+1].dis - sta[i].dis){
				length += cmax*davg;
				break;
			}
			else if(sta[i+1].price <= sta[i].price){
				if(nowgas*davg >= sta[i+1].dis-sta[i].dis){
					length += sta[i+1].dis-sta[i].dis;
					nowgas -= (sta[i+1].dis-sta[i].dis)/davg;
				}
				else{
					length += sta[i+1].dis-sta[i].dis;
					cost += ((sta[i+1].dis-sta[i].dis)/davg - nowgas) * sta[i].price;
					nowgas = 0;
				}
			}
			else{
				int len = cmax*davg;
				j = i+1;
				int next = i+1;
				int min = sta[i+1].price;
				while(sta[j].dis - sta[i].dis <= len){
					if(min >= sta[j].price){
						next = j;
						min = sta[j].price;
					}
					if(sta[j].price <= sta[i].price)
					    break;
					j++;
				}
				if(sta[j].dis - sta[i].dis <= len){
					if(nowgas*davg < sta[j].dis - sta[i].dis){
						cost += (sta[j].dis - sta[i].dis - nowgas*davg) / davg * sta[i].price;
						nowgas = 0;
						length += sta[j].dis - sta[i].dis;
					}
					else{
						length += sta[j].dis - sta[i].dis;
						nowgas -= (sta[j].dis - sta[i].dis) / davg;
					}
					i = j-1;
				}
				else{
					j = next;
					cost += (cmax - nowgas) * sta[i].price;
					nowgas = cmax - (sta[j].dis - sta[i].dis) / davg;
					length += sta[j].dis - sta[i].dis;
					i = j-1;
				}
			}
		}
		if(i < n){
			printf("The maximum travel distance = %.2lf
",length);
		}
		else{
			printf("%.2lf
",cost);
		}
	}	
	return 0;
}