HDU Bone Collector

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

 15 101 2 3 4 55 4 3 2 1 

 

Sample Output

 14 

 

dp：一维数组：

#include<stdio.h>#include<algorithm>#include<iostream>#include<string.h>using namespace std;int dp[1100],a[1000],b[1000];int main(){    int n,v,t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&v);        for(int i=0; i<n; i++)            scanf("%d",&a[i]);        for(int i=0;i<n;i++)            scanf("%d",&b[i]);        memset(dp,0,sizeof(dp));        for(int i=0; i<n; i++)        {            for(int j=v;j>=b[i];j--)            {                if(dp[j]<dp[j-b[i]]+a[i])//dp之后均更新成最优值，上次更新时某个剩余空间（〉=b[i]）的价值+a[i]的价值，可能比上次dp[j]的值大。                    dp[j]=dp[j-b[i]]+a[i];            }        }        cout<<dp[v]<<endl;    }    return 0;}

二维数组：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;int max(int a,int b){    return a>b?a:b;}int t,cost[1100],value[1100],dp[1100][1100];int main(){    scanf("%d",&t);    while(t--)    {        int n,v;        scanf("%d%d",&n,&v);        for(int i=1;i<=n;i++)            scanf("%d",&value[i]);        for(int i=1;i<=n;i++)            scanf("%d",&cost[i]);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            for(int j=0;j<=v;j++)            {                if(j>=cost[i])                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-cost[i]]+value[i]);                else                    dp[i][j]=dp[i-1][j];            }        }        cout<<dp[n][v]<<endl;    }    return 0;}