HDU 1114

 

http://acm.hdu.edu.cn/showproblem.php?pid=1114

完全背包问题

两个代码 内外循环次序互换了

代码一：

46MS

228K

724 B

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
const int inf=6000005;
using namespace std;
int dp[10003],w[505],v[505];
int main()
{
  int CASE,E,F,empty,n,i,j;
  scanf("%d",&CASE);
  while(CASE--)
  {
    scanf("%d%d%d",&E,&F,&n);
    empty=F-E;
    for(i=1;i<=n;i++)
    scanf("%d%d",&v[i],&w[i]);
    for(i=1;i<=empty;i++)
    dp[i]=inf;
    dp[0]=0;

     for(i=1;i<=n;i++)     //外层循环为种类
      for(j=w[i];j<=empty;j++)
      {
         //if(j>=w[i])
         dp[j]=min(dp[j],dp[j-w[i]]+v[i]);
      }

     if(dp[empty]==inf)
     printf("This is impossible.\n");
     else
     printf("The minimum amount of money in the piggy-bank is %d.\n",dp[empty]);

  }
  return 0;
}

代码2

125MS

228K

713 B

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
const int inf=6000005;
using namespace std;
int dp[10003],w[505],v[505];
int main()
{
  int CASE,E,F,empty,n,i,j;
  scanf("%d",&CASE);
  while(CASE--)
  {
    scanf("%d%d%d",&E,&F,&n);
    empty=F-E;
    for(i=1;i<=n;i++)
    scanf("%d%d",&v[i],&w[i]);
    for(i=1;i<=empty;i++)
    dp[i]=inf;
    dp[0]=0;
    for(j=0;j<=empty;j++)  //外侧循环为容量
     for(i=1;i<=n;i++)
     {
         if(j>=w[i])
         dp[j]=min(dp[j],dp[j-w[i]]+v[i]);
     }

     if(dp[empty]==inf)
     printf("This is impossible.\n");
     else
     printf("The minimum amount of money in the piggy-bank is %d.\n",dp[empty]);

  }
  return 0;
}