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 ?

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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.

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Output One integer per line representing the maximum of the total value (this number will be less than 2³¹).

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Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1

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Sample Output 14

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Author Teddy

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Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest 代碼(一維數(shù)組): 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 }

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代碼(二維數(shù)組):

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 }

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轉(zhuǎn)載于:https://www.cnblogs.com/ZERO-/p/9741042.html

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