Bone Collector

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 50442????Accepted Submission(s): 21153

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 #include<iostream> #include<stdio.h> using namespace std; const int maxx = 1005; int main(){int t;scanf("%d",&t);while(t--){int n,v;scanf("%d%d",&n,&v);int val[maxx];int wei[maxx];for(int i=1;i<=n;i++){scanf("%d",&val[i]);}for(int i=1;i<=n;i++){scanf("%d",&wei[i]);}int dp[maxx][maxx];for(int i=0;i<=v;i++)dp[0][i]=0;for(int i=1;i<=n;i++){for(int j=0;j<=v;j++){if(j<wei[i]){dp[i][j]=dp[i-1][j];}else{dp[i][j]=max(dp[i-1][j],dp[i-1][j-wei[i]]+val[i]);}}}printf("%d\n",dp[n][v]);} } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/superxuezhazha/p/5704210.html

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