HDU——2602Bone Collector（01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 46460    Accepted Submission(s): 19338

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

发现不做DP不行啊，以后还怎么混。先来道题入个门....

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
struct gutou
{
    int val;
    int weigh;
}bone[1010];
int dp[1010][1010];
int main (void)
{
    int t,i,n,j,v;
    cin>>t;
    while (t--)
    {
        memset(dp,0,sizeof(dp));
        cin>>n>>v;
        for (i=1; i<=n; i++)
        {
            cin>>bone[i].val;
        }
        for (i=1; i<=n; i++)
        {
            cin>>bone[i].weigh;
        }
        for (i=1; i<=n; i++)
        {
            for (j=v; j>=0; j--)
            {
            	if(j<bone[i].weigh)//二维的话必须要有这局
            	{
	            	dp[i][j]=dp[i-1][j];
	            	continue;
	            }
                dp[i][j]=max( dp[i-1][j] , dp[i-1][j-bone[i].weigh]+bone[i].val );//取和不取
            }
        }
        cout<<dp[n][v]<<endl;
    }
    return 0;
}