hdu 2602(dp)

题目链接 ：http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32499    Accepted Submission(s): 13379

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

 

 

 

题意 ：给你一个包的体积，每块骨头的价值和占用的体积，求出可以放入价值最大方案的价值。

分析 ：简单的01背包，纯属模板题，也是我做的第一题背包，就直接贴代码了。

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;

//dp[i][j]表示放入第i块骨头并占用j体积的最大价值，
//c[i]表示第i块骨头的体积
//w[i]表示第i块骨头的价值

int dp[1111][1111],c[1111],w[1111];
int T,N,V;

int main ()
{
    int i,j;
    scanf ("%d",&T);
    while (T--)
    {
        scanf ("%d%d",&N,&V);
        for (i=1; i<=N; i++)
        scanf ("%d",&w[i]);
        for (i=1; i<=N; i++)
        scanf ("%d",&c[i]);
        memset(dp, 0, sizeof(dp));
        for (i=1; i<=N; i++)
        {
            for (j=0; j<=V; j++)
            {
                dp[i][j] = dp[i-1][j];   //这里主要考虑j小于c[i]时放不下第i块骨头
                if (j >= c[i])
                dp[i][j] = max(dp[i-1][j], dp[i-1][j-c[i]]+w[i]);
            }
        }
        printf ("%d
",dp[N][V]);
    }
    return 0;
}