dp or 贪心 --- hdu : Road Trip

Road Trip

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 29, Accepted users: 29

Problem 12882 : No special judgement

Problem description

You are planning a road trip to visit your friends, each of whom live in different towns. Of course, you don't want to pay any more for fuel on the trip than you need to. However, the price of fuel in each of the towns is different, so you will need to carefully plan the trip to minimise the cost. For example, it might make sense to take on extra fuel at a town where the price is low so that you don't need to buy so much at a town where it is more expensive. Indeed, it may even make sense to sell excess fuel at some towns to recoup some of the costs. Of course, your car can only hold a certain amount of fuel, and you have to make sure you take on enough fuel at each town to reach the next (assume that it's OK to arrive with an empty tank). Your task is to write a program to help you plan your trip.

Input

Input will consist of specifications for a series of journeys. Information for each journey will begin with a line containing an integer 0 < c < 100 that specifies the capacity of the car's fuel tank in litres and an integer 0 < t < 20 that specifies the number of towns to visit for that journey. A line containing two zeros indicates the end of input. The following t lines contain information about successive stages on the journey: the price (in fixed point dollars-and-cents format, 0.01 <= p < 9.99) of one litre of fuel (either to buy or to sell) in the town at the beginning of the stage, and the number of litres of fuel (an integer, 1 <= n < 100) needed to reach the next town.

Output

Output should consist of one line for each journey comprising the journey number (formatted as shown) followed by a single space and the minimum cost of completing that journey, formatted as a fixed-point number with 2 decimal places.

Sample Input

10 3 2.00 7 1.50 8 1.00 3 50 6 1.50 20 4.20 5 1.15 35 1.41 27 1.92 30 2.21 15 0 0

Sample Output

Journey 1: 29.00 Journey 2: 117.64

Problem Source

HNU Contest

---

 

Mean:

你要开车去n个小镇拜访你的老朋友，但是你不是高富帅，只得将油钱尽量的省，现在给你油箱的容量、小镇数量、每个小镇的油价、到达下一个小镇需要的油，你可以买油也可卖油，问你最少的油钱是多少。

analyse:

这题就是一个贪心的思想，如果下一站的油价比这站高，那么别想了，将油箱装满，到下一站就可以赚钱；反之就只需买够路上用的就可。

当然有人说这题是dp，这也没错，看个人怎么理解，其实这题就两个状态之间转移，dp已经退化为贪心。

Time complexity：O(n)

 

Source code：

//Memory   Time
// 1254K    0MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 110
#define LL long long
using namespace std;
struct Node
{
   double p;
   int next;
};
Node node[MAX];
int v,n;
int main()
{
   int kase=1;
   while(scanf("%d %d",&v,&n),v+n)
   {
       double cost=0.0;
       printf("Journey %d: ",kase++);
       bool flag=0;
       for(int i=1;i<=n;i++)
       {
           scanf("%lf %d",&node[i].p,&node[i].next);
           if(node[i].next>v)
           {
               flag=1;
               break;
           }
       }
       if(flag==1)
       {
           puts("0.00");
           continue;
       }
       int now=0;
       for(int i=1;i<n;i++)   //zui hou mei pan
       {


            if(i>1) // begin to No.2
           {
               if(node[i].p>node[i-1].p&&now>node[i].next)   //sell
               {
                   int tmp=now-node[i].next;
                   cost-=node[i].p*tmp;
                   now-=tmp;
               }
           }


           if(node[i].p<node[i+1].p)   //  earn money
           {
               int tmp=v-now;
               cost+=node[i].p*tmp;
               now=v-node[i].next;
           }
           else    // bu ke zhuan qian
           {
               int tmp;
               if(node[i].next>now)
               {
                   tmp=node[i].next-now;
                   now=0;
               }
               else
               {
                   tmp=0;
                   now-=node[i].next;
               }
               cost+=node[i].p*tmp;
           }

       }
       // final
       if(now>node[n].next)
       {
           int tmp=now-node[n].next;
           cost-=tmp*node[n].p;
           now=0;
       }
       else
       {
           int tmp=node[n].next-now;
           cost+=tmp*node[n].p;
           now=0;
       }

       printf("%.2lf\n",cost);
   }
   return 0;
}