A - Bone Collector——HOJ 简单0-1背包

A - Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave … 
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ? 

 

Input

The first line contain a integer T , the number of cases. 
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the maximum of the total value (this number will be less than 2  ³¹).

 

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

 

Sample Output

14

 

#include<stdio.h>
#include<string.h>
#include<math.h>
long my_max(long x,long y){return x>y?x:y;}

long value[1005],weight[1005];
long dp[1005][1005];
int main()
{
	int T,i,j,n,v;
	scanf("%d",&T);
	while(T--)
	{
		memset(dp,0,sizeof(dp));
		memset(value,0,sizeof(value));
		memset(weight,0,sizeof(weight));
		scanf("%d %d",&n,&v);
		for(i=1;i<=n;i++)
			scanf("%ld",&value[i]);
		for(i=1;i<=n;i++)
			scanf("%ld",&weight[i]);
		
		for(i=1;i<=n;i++)
		{
			for(j=0;j<=v;j++)//j是从0开始，不是从1开始 
			{
				dp[i][j]=dp[i-1][j];
				if(weight[i]<=j)
				{
					dp[i][j]=my_max(dp[i][j],dp[i-1][j-weight[i]]+value[i]);
				}
			}
		}
		printf("%ld\n",dp[n][v]);
	}
	return 0;
}