acm 3 1017 背包问题

1.1017

2.Problem Q 
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 152   Accepted Submission(s) : 57
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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>
 


Input
The first line contain a integer T , the number of cases.<br>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

3.背包问题

4.骨头有w容量和v价值，问容量中最大能搜集多少价值的骨头（真是独特的兴趣-  -）

5.#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
    int T,n,Vol,i,j;
    int a;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&Vol);
        int val[1005];
        int vol[1005];
        int dp[1005];
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
        }
        for(i=1;i<=n;i++)
        {
            scanf("%d",&vol[i]);
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            for(j=Vol;j>=vol[i];j--)
            {
                dp[j]=max(dp[j],dp[j-vol[i]]+val[i]);
            }
        
        }
        cout<<dp[Vol]<<endl;
    }
}