HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23264    Accepted Submission(s): 9443

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

 

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典型的01背包题目，说实话我是看了别人的算法模板的，还不是很理解，明天接着刷类似的题目吧，期待会有所领会。

代码如下：

#include <stdio.h>#include <string.h>#define rep(i,k,n) for(int i=(k); i<(n); i++)#define cls(x,a) memset(x,a,sizeof(x))#define maxn 10010int a[maxn],b[maxn];int main(void){    int n,m,t;    int dp[maxn];    scanf("%d",&t);    rep(i,0,t){     scanf("%d%d",&n,&m);     rep(i,0,n) scanf("%d",&a[i]);     rep(i,0,n) scanf("%d",&b[i]);     cls(dp,0);     rep(i,0,n)      for(int j=m; j>=b[i]; j--){       if (dp[j]<dp[j-b[i]]+a[i])        dp[j] = dp[j-b[i]]+a[i];      }     printf("%d\n",dp[m]);    }}