HDU 2602 ——背包问题

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


题目大意：有n堆物品，物品有各自的价值和重量，问怎样在装满背包的时候使得背包内的物品价值最高。

解题思路：一开始的时候我是想到的排序方法，用2维数组去求，但是当我查了下网上资料后，发现还有另外一种方法
，就是状态转移：f[j]=max{f[j],f[j-b[i].volume]+b[i].value}

具体方法可以在代码里面看：

#include <cstdio>
#include <cstring>
#include <iostream>


using namespace std;


struct things
{
    int h;
    int v;
}goods[1005];

int max(int a,int b)
{
    return a>b?a:b;
}


int main()
{
    int t;
    int n,v,l;
    int dp[1005];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&v);
        int i;
        for(i = 1;i<=n;i++)
            scanf("%d",&goods[i].h);
        for(i = 1;i<=n;i++)
            scanf("%d",&goods[i].v);
        memset(dp,0,sizeof(dp));
        for(i = 1;i<=n;i++)
        {
            for(l = v;l>=goods[i].v;l--)
                dp[l] = max(dp[l],dp[l-goods[i].v]+goods[i].h);
        }

        printf("%d
",dp[v]);
    }

    return 0;
}