【HDU 2602】Bone Collector（01揹包）

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

題目大意

揹包問題，給你兩個數N，V分別表示骨頭的個數和揹包的容量，接下來一行表示每個骨頭的價值，在接下來的一行表示每個骨頭的重量。求最大能被價值爲多少的骨頭。

思路

01揹包裸體

代碼

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

using namespace std;
const int maxn=1000+5;

int w[maxn],v[maxn],dp[maxn],t,n,cost;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&cost);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=cost;j>=w[i];j--)
            {
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[cost]);
    }
    return 0;
}