Bone Collector

最新推荐文章于 2021-01-18 14:47:36 发布

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本文介绍了一种经典的背包问题——骨收集者问题。通过给定的案例，文章详细阐述了如何计算骨收集者在限定体积内能获得的最大总价值。采用动态规划方法，通过示例代码展示了问题解决的具体步骤。

Bone Collector
Time Limit: 1000MS Memory Limit: 65536KB
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).

Example Input

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

Example Output

Hint
hdoj2602 http://blog.youkuaiyun.com/u014231159/article/details/25605411?utm_source=tuicool&utm_medium=referral
Author

think:
这又是一道英文题，其实自己不太喜欢做英文题，因为有的地方读不懂的感觉实在是不好受，但我们都需要挑战自己，为将来打下基础。
回归题目，这是一道背包问题，大意就是给出了几组测试例子，然后又给出了骨头的数量和骨收集器的体积，然后接着是两行，第一行包含每个骨头的价值，第二行包含每个骨头的体积，要求计算出骨收集器的最大总价值。
代码实现：

 #include<stdio.h>
int max(int a, int b)
{
   if(a>b)
   return a;
   else
   return b;
}
int main()
{
    int x, n;
    int p[100005], m[100050];
    int sum[200000];
    int t;
    scanf("%d", &t);
    while(t--)
    {
    scanf("%d %d", &n, &x);
       int i, j;
       memset(sum, 0, sizeof(sum));
       for(i=0;i<n;i++)
       {
         scanf("%d", &m[i]);
       }
         for(i=0;i<n;i++)
       {
         scanf("%d", &p[i]);
       }
       for(i=0;i<n;i++)
       {
         for(j=x;j>=0;j--)
         {
            if(j>=p[i])
            sum[j] = max(sum[j-p[i]]+m[i], sum[j]);
         }
       }
     printf("%d\n", sum[x]);
    }
    return 0;
}