PAT 1033. To Fill or Not to Fill

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1033

题解：

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<algorithm>
using namespace std;
#define INF 0x6fffffff
struct point
{
    double price;
    double distance;
} node[505];
bool cmp(const struct point &a,const struct point &b)
{
    return a.distance<b.distance;
}
int main()
{
    double cmax,davg,d,nowCapacity,minPrice,summ;
    int n,idx;
    double len;//满邮箱最长行驶距离
    bool flag;
    scanf("%lf%lf%lf%d",&cmax,&d,&davg,&n);
    len=cmax*davg;
    for(int i=0; i<n; ++i)
        scanf("%lf%lf",&node[i].price,&node[i].distance);
    sort(node,node+n,cmp);
    node[n].price=0;
    node[n].distance=d;
    if(node[0].distance>0)
    {
        printf("The maximum travel distance = 0.00\n");
    }
    else
    {
        flag=true;
        nowCapacity=0.0;
        summ=0.0;
        for(int i=0; i<n;)
        {
            if(node[i+1].distance-node[i].distance>len)//两站距离大于最大行驶距离
            {
                flag=false;
                printf("The maximum travel distance = %.2f\n",node[i].distance+ len);
                break;
            }
            minPrice=node[i].price;
            idx=i;
            for(int j=i+1; j<=n&&node[j].distance-node[i].distance<=nowCapacity*davg; ++j)
            {//找出当前油箱里的油能到达的所有加油站里，油价最便宜的那个
                if(node[j].price<minPrice)
                {
                    minPrice=node[j].price;
                    idx=j;
                }
            }
            if(idx!=i)
            {
                nowCapacity-=(node[idx].distance-node[i].distance)/davg;
                i=idx;
            }
            else
            {//若找不到，找出最近的一个能到达的比当前油价便宜的站，加一些油，跑到那个站
                idx=i;
                for(int j=i+1; j<=n&&node[j].distance-node[i].distance<=len; ++j)
                {
                    if(node[j].price<node[i].price)
                    {
                        idx=j;
                        break;
                    }
                }
                if(idx!=i)
                {
                    summ+=((node[idx].distance-node[i].distance)/davg-nowCapacity)*node[i].price;
                    nowCapacity=0;
                    i=idx;
                }
                else
                {   //找不到比当前油站的价格还便宜的油站的时候
                    //在当前油站需要加满油，跑到能跑到的所有站里油价最小的那个油站
                    idx=i;
                    minPrice=INF;
                    for(int j=i+1; j<=n&&node[j].distance-node[i].distance<=len; ++j)
                    {
                        if(node[j].price<minPrice)
                        {
                            minPrice=node[j].price;
                            idx=j;
                        }
                    }
                    summ+=(cmax-nowCapacity)*node[i].price;
                    nowCapacity=cmax-(node[idx].distance-node[i].distance)/davg;
                    i=idx;
                }
            }
        }
        if(flag)
            printf("%.2f\n",summ);
    }
    return 0;
}

来源： http://blog.csdn.net/acm_ted/article/details/20592059