HDU - 2602 Bone Collector-白红宇

HDU - 2602 Bone Collector

阅读量：4338 次

发布时间：2019-06-07

本文共 1816 字，大约阅读时间需要 6 分钟。

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 2

³¹).

Sample Input15 101 2 3 4 55 4 3 2 1Sample Output14

********************************************

分析：背包模型

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
            8 #include 
            
              9 using namespace std;10 11 #define N 250012 #define INF 0x3f3f3f3f13 14 int p[N],v[N],dp[25000];15 16 int main()17 {18 int T,n,m,i,j;19 20 scanf("%d", &T);21 22 while(T--)23 {24 memset(dp,0,sizeof(dp));25 26 scanf("%d %d",&n, &m);27 28 for(i=1;i<=n;i++)29 scanf("%d", &p[i]);30 for(i=1;i<=n;i++)31 scanf("%d", &v[i]);32 33 for(i=1;i<=n;i++)34 for(j=m;j>=v[i];j--)35 dp[j]=max(dp[j],dp[j-v[i]]+p[i]);36 37 printf("%d\n", dp[m]);38 }39 return 0;40 }

转载于:https://www.cnblogs.com/weiyuan/p/5775306.html

你可能感兴趣的文章