HDU-2602-Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 45832 Accepted Submission(s): 19075

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

01背包 动规基础题
状态转移方程：dp[j]=max(dp[j],dp[j-num[i].体积]+num[i].价值
代码

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string>
#include<string.h>
#include<stdio.h>
#include<queue>
using namespace std;
//01背包
struct node
{
    int value;//价值
    int volume;//体积
} num[1005];
int dp[1005];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        int N;//数量
        int max_v;//背包最大体积
        scanf("%d%d",&N,&max_v);
        for(int i=0; i<N; i++)
            scanf("%d",&num[i].value);
        for(int i=0; i<N; i++)
            scanf("%d",&num[i].volume);
        for(int i=0; i<N; i++)
        {
            for(int j=max_v; j>=num[i].volume; j--)
            {
                dp[j]=max(dp[j],dp[j-num[i].volume]+num[i].value);
            }
        }
        printf("%d\n",dp[max_v]);
    }
    return 0;
}

背包 背包 背包