hdu-2602-Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40361    Accepted Submission(s): 16764

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

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

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1203 2159 2955 1171 2191 

 

题目分析：

就是简单的背包，但是要读清题，第一行是 石头数（也就是价值）  第二行是 需要的背包体积。但是一定要搞清楚 石头数可以是 0  背包体积也可以是 0.而且石头数（就是价值）可以不是0 ，但消耗的体积可以0

5  0

1 2 3 4 5

0 0 0 1 0

结果是： 11  

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int f[1010][1010];int cost[1010],val[1010];int main(){int t;scanf("%d",&t);while(t--){int n,v;scanf("%d%d",&n,&v);memset(f,0,sizeof(f));cost[0]=val[0]=0;for(int i=1;i<=n;i++)    scanf("%d",&val[i]);for(int i=1;i<=n;i++)    scanf("%d",&cost[i]);for(int i=1;i<=n;i++)for(int j=0;j<=v;j++)  // 记住要从0开始  石头是可以不消耗体积的{f[i][j]=f[i-1][j];if(j>=cost[i])f[i][j]=max(f[i-1][j],f[i-1][j-cost[i]]+val[i]);}printf("%d\n",f[n][v]);}return 0;}

一维数组优化代码

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int f[1010];int cost[1010],val[1010];int main(){int t;scanf("%d",&t);while(t--){int n,v;scanf("%d%d",&n,&v);memset(f,0,sizeof(f));cost[0]=val[0]=0;for(int i=1;i<=n;i++)    scanf("%d",&val[i]);for(int i=1;i<=n;i++)    scanf("%d",&cost[i]);for(int i=1;i<=n;i++)for(int j=v;j>=0;j--)//  记者要到 0 {if(j>=cost[i])f[j]=max(f[j],f[j-cost[i]]+val[i]);}printf("%d\n",f[v]);}return 0;}