hdu 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 30095 Accepted Submission(s): 12389



Problem 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 231).


Sample Input

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

Sample Output

14/*题解：（对01背包问题的个人理解） 
     01背包问题，0代表不放，1代表放，即是对第i个物品放或者不放的背包问题。
     01背包问题通常具有以下特征：
        1.物品有很多种，但是每种仅有一个
        2.背包容量有限，题目要求得到最优解
        3.通过递推的方式，可以求得最优解 
*/
[code]#include<cstdio>
#include<cstring>
int dp[1010][1010];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int T,n,v,i,j,w[1010],val[1010];
    scanf("%d",&T);
    while(T--)
	{
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&v);
        for(i=1; i<=n; i++)
            scanf("%d",&val[i]);
        for(i=1; i<=n; i++)
            scanf("%d",&w[i]);
        for(i=1; i<=n; i++)
		{
            for(j=0; j<=v; j++)
			{
                if(j>=w[i])
				{        //放还是不放 
                     dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+val[i]);
                }
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        printf("%d\n",dp
[v]);
    }
    return 0;
}

/*题解：
滚动数组（一维数组），解01背包问题。
参考自《算法竞赛入门经典第二版》p273
*/
#include<cstdio>
#include<cstring>
int dp[1010];
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int T,n,v,i,j,w[1010],val[1010];
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d %d",&n,&v);
for(i=1; i<=n; i++)
scanf("%d",&val[i]);
for(i=1; i<=n; i++)
scanf("%d",&w[i]);
for(i=1; i<=n; i++)
{
for(j=v; j>=0; j--)
{
if(j>=w[i]) dp[j]=max(dp[j],dp[j-w[i]]+val[i]);
}
}
printf("%d\n",dp[v]);
}
return 0;
}
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