Bone Collector

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

Total Submission(s): 11188    Accepted Submission(s): 4317

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 2

³¹).

 

 

 

1 /*   2 //0-1背包问题，容量可以从前向后推（二维数组），也可以从后向前推（一维数组）  3  4 //代码一： 5 #include
      
        6 #include
       
         7 int main() 8 { 9     int t,V,i,j,N;10     int val[1001],vol[1001],record[1001];11     scanf("%d",&t);12     while(t--)13     {14         memset(record,0,sizeof(record));15         scanf("%d%d",&N,&V);16         for(i=0;i
        
         =vol[i];--j)22                 if(record[j-vol[i]]+val[i]>record[j])23                     record[j]=record[j-vol[i]]+val[i];24         printf("%d\n",record[V]);25     }26     return 0;27 }28 29 30 //代码二：31 #include
         
          32 #include
          
           33 int main()34 {35     int t,V,i,j,N;36     int val[1001],vol[1001],record[1001];37     scanf("%d",&t);38     while(t--)39     {40         memset(record,0,sizeof(record));41         scanf("%d%d",&N,&V);42         for(i=0;i
           
            =vol[i];--j)48 if(record[j-vol[i]]+val[i]>record[j])49 record[j]=record[j-vol[i]]+val[i];50 }51 printf("%d\n",record[V]);52 }53 return 0;54 }55 */56 57 //代码三： 二位数组的写法58 #include
            
             59 #include
             
              60 61 int val[1001],vol[1001];62 int dp[1001][1001];63 int main()64 {65 int T,i,j,n,v;66 scanf("%d",&T);67 while(T--)68 {69 scanf("%d%d",&n,&v);70 for(i=1;i<=n;++i)71 scanf("%d",&val[i]);72 for(i=1;i<=n;++i)73 scanf("%d",&vol[i]);74 memset(dp,0,sizeof(dp));75 for(i=1;i<=n;++i)76 for(j=0;j<=v;++j) //注意这里要从0开始77 if(j>=vol[i]) //背包容量能过装下第i件骨头的情况下选取装与不装的最大值78 dp[i][j]=dp[i-1][j]>dp[i-1][j-vol[i]]+val[i]?dp[i-1][j]:dp[i-1][j-vol[i]]+val[i];79 else //背包容量不能装下第i件的化当前最大值为装下前i-1件骨头的状态80 dp[i][j]=dp[i-1][j];81 printf("%d\n",dp[n][v]);82 }83 return 0;84 }