hdu 2602 Bone Collector

网址：点击打开链接

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 61757    Accepted Submission(s): 25729

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

就是用动态规划，推荐一个讲的不错的博客，看懂他在表达什么意思了，然后自己推一遍，然后这个题迎刃而解。下面是我的代码！

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1005];int b[1005];int dp[1005][1005];int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(dp,0,sizeof(dp));        int i,n,v,j;        scanf("%d%d",&n,&v);        for(i=1; i<=n; i++)            scanf("%d",&a[i]);        for(i=1; i<=n; i++)            scanf("%d",&b[i]);        for(i=0; i<=v; i++)        {            if(i<b[n])                dp[n][i]=0;            else                dp[n][i]=a[n];        }        for(i=n-1; i>=1; i--)        {            for(j=0; j<=v; j++)            {                if(j<b[i])                    dp[i][j]=dp[i+1][j];                else                    dp[i][j]=max(dp[i+1][j],dp[i+1][j-b[i]]+a[i]);            }        }        printf("%d\n",dp[1][v]);        /* for(i=0;i<=n;i++)         {             for(j=0;j<=v;j++)                 printf("%d ",dp[i][j]);             printf("\n");         }*/    }    return 0;}