Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 32408 Accepted Submission(s): 13329

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

分析：

http://blog.csdn.net/dapengbusi/article/details/7463968

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 int v[1000],w[1005];
 6 int n,c;
 7 int f[1005][1005];//对于dp(i,j)就表示可选物品为i…n背包容量为j(总重量)时背包中所放物品的最大价值。
 8 void backpack()
 9 {
10     int i,j;
11     for(i=1;i<=n;i++)
12       for(j=0;j<=c;j++)
13       {
14           if(i==1)
15              f[i][j]=0;
16           else
17             f[i][j]=f[i-1][j];
18          if(j>=v[i])
19            f[i][j]=max(f[i][j],f[i-1][j-v[i]]+w[i]);
20       }
21     printf("%d
",f[n][c]);
22 }
23 int main()
24 {
25    int t;
26    scanf("%d",&t);
27   while(t--)
28   {
29       int i;
30      scanf("%d%d",&n,&c);
31     for(i=1;i<=n;i++)
32       scanf("%d",&w[i]);
33     for(i=1;i<=n;i++)
34       scanf("%d",&v[i]);
35     backpack();
36   }
37  return 0;
38 }