HDU2602Bone Collector 簡單0-1背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 48618????Accepted Submission(s): 20269

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

注意：先輸入每個骨頭的價值然后輸入每個骨頭的體積。

狀態轉移方程：dp[j]=max(dp[j],dp[j-v[i]]+w[i]。

#include <iostream> #include <string.h>using namespace std; int N,V; int dp[1111]; int w[1111],v[1111]; int main() {int t;while(cin>>t){while(t--){cin>>N>>V;for(int i=1;i<=N;i++){cin>>w[i];}for(int i=1;i<=N;i++){cin>>v[i];}memset(dp,0,sizeof(dp));for(int i=1;i<=N;i++){for(int j=V;j>=v[i];j--){ //j如果小于了v[i]那么v[i]一定無法裝入袋子dp[j]=max(dp[j],dp[j-v[i]]+w[i]);}}cout<<dp[V]<<endl;}}return 0; }

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posted on 2016-06-14 21:44 asuml 閱讀(...) 評論(...) 編輯 收藏

轉載于:https://www.cnblogs.com/asuml/p/5585570.html

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