HDU 2602 Bone Collector DP(01背包)

Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

很基础的一道dp熟悉一下

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;

const int MAXN=1005;
int dp[MAXN][MAXN];
int value[MAXN];
int volume[MAXN];

int main()
{
    //FIN
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,v;
        scanf("%d%d",&n,&v);
        for(int i=1;i<=n;i++)
            scanf("%d",&value[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&volume[i]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            for(int j=0;j<=v;j++)
            {
                if(j>=volume[i])
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);
                else
                    dp[i][j]=dp[i-1][j];
            }
        printf("%d\n",dp[n][v]);
    }

}