苏州大学新生寒假训练day3 D - Bone Collector

Problem:

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:

Sample Output:

很明显的01背包问题，默认了每一个骨头都是独一无二的，注意数据的多组数据即可，01背包的矩阵dp问题在我的其它blog里也有的。最近在学dp多刷一刷。

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

int main()
{
    long long n;
    cin>>n;

    while(n--)
    {
        long long a;
        long long size;

        cin>>a>>size;
        long long v[100000];
        long long w[100000];
        long long dp[100000];
        memset(v,0,sizeof(v));
        memset(w,0,sizeof(w));
        memset(dp,0,sizeof(dp));
        for(int i = 1;i <= a;i++)
        {
            cin>>v[i];
        }
        for(int i = 1;i <= a;i++)
        {
            cin>>w[i];
        }
        for(int i =1;i <= a;i++)
        {
            for(int j = size;j >= w[i];j--)
            {
                dp[j] = max(dp[j],dp[j-w[i]] + v[i]);
            }
        }
        cout<<dp[size]<<endl;
    }
    return 0;
}