HDUOJ 2602 Bone Collector

题目链接

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背包裸题~
我们用 $dp[j]$dp[j] 表示容量为 $j$j 时获得的最大价值，则有状态转移方程：
$dp[j]=max(dp[j],dp[j-w[i]]+v[i]),i\in[0,n)$dp[j]=max(dp[j],dp[j−w[i]]+v[i]),i∈[0,n)
AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

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