hdoj-2602Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 67533    Accepted Submission(s): 28197

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

题目链接

01背包模板题，下面两个模板>_>

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int dp[1005][1005];int main(){    int i,j,m,n;    int t;    int w[1005];    int v[1005];    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&m,&n);        for(i=1;i<=m;i++)            scanf("%d",&w[i]);        for(i=1;i<=m;i++)            scanf("%d",&v[i]);        memset(dp,0,sizeof(dp));        for(i=1;i<=m;i++)        {            for(j=0;j<=n;j++)            {                if(v[i]<=j)                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);                else                    dp[i][j]=dp[i-1][j];            }        }        printf("%d\n",dp[m][n]);    }    return 0;}

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int dp[1005];int main(){    int i,j,m,n;    int t;    int w[1005];    int v[1005];    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&m,&n);        for(i=1;i<=m;i++)            scanf("%d",&w[i]);        for(i=1;i<=m;i++)            scanf("%d",&v[i]);        memset(dp,0,sizeof(dp));        for(i=1;i<=m;i++)        {            for(j=n;j>=v[i];j--)            {                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);            }        }        printf("%d\n",dp[n]);    }    return 0;}