YT06-揹包-1001—Bone Collector -（6.27日-煙臺大學ACM預備隊解題報告）

kewlgrl發表於2015-07-15
ACM

Bone Collector

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 58   Accepted Submission(s) : 25

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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

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

Author

Teddy

Source

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


題意：用包按順序收集骨頭，給出包可收納骨頭的數量和各種骨頭的數量及其價值，求可獲得的最大價值。
思路：每種骨頭按照包的剩餘容量選擇是否收集
狀態：large[j]:表示收集了volume為j時的最大價值
狀態轉移：large[j]=max(large[j],large[j-volume[i]]+value[i])


#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
  int i,j;
  int t,n,v;
  int large[1006],volume[1006],value[1006];
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d%d",&n,&v);
  for(i=1;i<=n;i++)
      scanf("%d",&value[i]);
    for(i=1;i<=n;i++)     scanf("%d",&volume[i]);memset(large,0,sizeof(large));
    for(i=1;i<=n;i++)
    for(j=v;j>=volume[i];j--)  large[j]=max(large[j],large[j-volume[i]]+value[i]);
    printf("%d\n",large[v]);
  }
  return 0;
}


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