我快被逼疯了！MixingMilk

我真的快被这个题逼疯了，提交总是time limit exceed。我已经改了N次了，我觉得已经很快了啊~~~~

Mixing Milk

 Source : Unknown Time limit : 3 sec Memory limit : 32 M

Submitted : 7091, Accepted : 2949

Since milk packaging is such a low margin business, it is important to keep the price of the raw product (milk) as low as possible. Help Merry Milk Makers get the milk they need in the cheapest possible manner.

The Merry Milk Makers company has several farmers from which they may buy milk, and each one has a (potentially) different price at which they sell to the milk packing plant. Moreover, as a cow can only produce so much milk a day, the farmers only have so much milk to sell per day. Each day, Merry Milk Makers can purchase an integral amount of milk from each farmer, less than or equal to the farmer's limit.

Given the Merry Milk Makers' daily requirement of milk, along with the cost per gallon and amount of available milk for each farmer, calculate the minimum amount of money that it takes to fulfill the Merry Milk Makers' requirements.

Note: The total milk produced per day by the farmers will be sufficient to meet the demands of the Merry Milk Makers.

Input

The first line contains two integers, N and M. The first value, N, (0 <= N <= 2,000,000) is the amount of milk that Merry Milk Makers' want per day. The second, M, (0 <= M <= 5,000) is the number of farmers that they may buy from.

The next M lines (Line 2 through M+1) each contain two integers, Pi and Ai. Pi (0 <= Pi <= 1,000) is price in cents that farmer i charges. Ai (0 <= Ai <= 2,000,000) is the amount of milk that farmer i can sell to Merry Milk Makers per day.

Output

A single line with a single integer that is the minimum price that Merry Milk Makers can get their milk at for one day.

Sample Input

100 5 5 20 9 40 3 10 8 80 6 30

Sample Output

630

 

最初

#include <stdio.h>
#include <stdlib.h>
/*time limit exceed*/
int main()
{
    int n,m,p[5000],a[5000],i,j,temp,price;
    while(1)
    {
        scanf("%d",&n);
        scanf("%d",&m);
        for(i=0;i<m;i++)
        {
             scanf("%d %d",&p[i],&a[i]);  
        }
        for(i=0;i<m;i++)
        {
            for(j=i+1;j<m;j++)
            {
                if(p[i]>p[j])
                {
                     temp=p[j];
                     p[j]=p[i];
                     p[i]=temp;
                    
                     temp=a[j];
                     a[j]=a[i];
                     a[i]=temp;            
                }                 
            }
            if(n>a[i])
            {
                price+=p[i]*a[i];
                n=n-a[i];
            }
            else if(n<a[i])
            {
                price+=p[i]*n;
                break;
            }
        }     
        printf("%d",price);
    }
    system("pause"); 
    return 0;
}

 

改后

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n,m,p[5000],a[5000],i,j,pp,aa,tp=-1,price=0;
    while(1)
    {
        scanf("%d",&n);
        scanf("%d",&m);
        for(i=0;i<m;i++)
             scanf("%d %d",&p[i],&a[i]);  
        while(n)
        {
            pp=1000;
            aa=0;
            for(j=0;j<m;j++)
            {
                if(tp==p[j])
                {
                    p[j]=a[j]=1001;
                    continue;
                }
                if(pp>p[j])
                {
                     pp=p[j];
                     aa=a[j];
                }            
            }
            if(n>=aa)
            {
                price+=pp*aa;
                n=n-aa;
            }
            else
            {
                price+=pp*n;
                n=0;
            }
            tp=pp;
        }     
        printf("%d\n",price);
    }
    system("pause"); 
    return 0;
}

 

再改后

#include <stdio.h>

int main()
{
     int n,m,p[5000],a[5000],i,j,temp,price;
     while(1)
     {
        price=0;
        scanf("%d",&n);
        scanf("%d",&m);
        for(i=0;i<m;i++)
             scanf("%d %d",&p[i],&a[i]);  
        i=0;
        while(n)
        {
            for(j=i+1;j<m;j++)
            {
                if(p[i]>p[j])
                {
                     temp=p[j];
                     p[j]=p[i];
                     p[i]=temp;
                     temp=a[j];
                     a[j]=a[i];
                     a[i]=temp;              
                }          
            }
            if(n>=a[i])
            {
                price+=p[i]*a[i];
                n=n-a[i];
            }
            else
            {
                price+=p[i]*n;
                n=0;
            }
            i++;
        }     
        printf("%d\n",price);
      }
      return 0;
}

 

我已经在不断的去改进了。为什么还是不行，谁能告诉我一个不超时的方法啊~~