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

Author

Teddy

Source

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

代码(一维数组):

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e5+10;
 4 int val[N],wei[N],dp[N];
 5 int main(){
 6     int t,n,m;
 7     scanf("%d",&t);
 8     while(t--){
 9             memset(val,0,sizeof(val));
10             memset(wei,0,sizeof(wei));
11             memset(dp,0,sizeof(dp));
12         scanf("%d%d",&n,&m);
13         for(int i=0;i<n;i++)
14             scanf("%d",&val[i]);
15         for(int i=0;i<n;i++)
16             scanf("%d",&wei[i]);
17         for(int i=0;i<n;i++){
18             for(int j=m;j>=wei[i];j--){
19                 dp[j]=max(dp[j],dp[j-wei[i]]+val[i]);
20             }
21         }
22         printf("%d\n",dp[m]);
23     }
24     return 0;
25 }

代码(二维数组):

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e3+10;
 4 int val[N],wei[N],dp[N][N];
 5 int main(){
 6     int t,n,m;
 7     scanf("%d",&t);
 8     while(t--){
 9         memset(dp,0,sizeof(dp));
10         scanf("%d%d",&n,&m);
11         for(int i=1;i<=n;i++)
12             scanf("%d",&val[i]);
13         for(int i=1;i<=n;i++)
14             scanf("%d",&wei[i]);
15         for(int i=1;i<=n;i++){
16             for(int j=0;j<=m;j++){
17                 if(wei[i]<=j)dp[i][j]=max(dp[i-1][j],dp[i-1][j-wei[i]]+val[i]);
18                 else dp[i][j]=dp[i-1][j];
19             }
20         }
21         printf("%d\n",dp[n][m]);
22     }
23     return 0;
24 }

转载于:https://www.cnblogs.com/ZERO-/p/9741042.html

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