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\n");
            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\n",length);
        }
        else{
            printf("%.2lf\n",cost);
        }
    }   
    return 0;
}
posted @ 2016-03-12 17:11 阿文的博客 阅读(...) 评论(...) 编辑 收藏