HDU 2602.Bone Collector-动态规划0-1背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 85530    Accepted Submission(s): 35381

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

 

 

Author

Teddy

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

 

代码(一维数组):

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int val[N],wei[N],dp[N];
int main(){
    int t,n,m;
    scanf("%d",&t);
    while(t--){
            memset(val,0,sizeof(val));
            memset(wei,0,sizeof(wei));
            memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",&val[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&wei[i]);
        for(int i=0;i<n;i++){
            for(int j=m;j>=wei[i];j--){
                dp[j]=max(dp[j],dp[j-wei[i]]+val[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}

 

代码(二维数组):

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
int val[N],wei[N],dp[N][N];
int main(){
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&val[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&wei[i]);
        for(int i=1;i<=n;i++){
            for(int j=0;j<=m;j++){
                if(wei[i]<=j)dp[i][j]=max(dp[i-1][j],dp[i-1][j-wei[i]]+val[i]);
                else dp[i][j]=dp[i-1][j];
            }
        }
        printf("%d\n",dp[n][m]);
    }
    return 0;
}