HDU 2602 Bone Collector（01背包）

最新推荐文章于 2021-01-18 14:47:36 发布

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本文详细解析了01背包问题的两种解决方案，一种使用二维数组，另一种采用滚动数组（一维）。通过具体示例展示了如何求解骨收集者在限定背包容量下获得的最大价值。

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Bone Collector

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

Sample Output

【思路分析】

裸的01背包问题。用了两种方法，第一种是用二维数组，第二种是用滚动数组（一维），两种方法都可以一边输入一边处理。滚动数组的具体操作见小白书。

代码如下：

1、二维数组

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int n,v;
int vol;
int val[maxn];
int dp[maxn][maxn];
int maxs(int a,int b)
{
    return a > b ? a : b;
}
void init()
{
   scanf("%d %d",&n,&v);
   for(int i = 1;i <= n;i++)
   {
      scanf("%d",&val[i]);
   }
   for(int i = 1;i <= n;i++)
   {
       scanf("%d",&vol);
       for(int j = 0;j <= v;j++)
       {
          dp[i][j] = (i == 1 ? 0 : dp[i - 1][j]);
          //dp[i][j]表示把前i个物品放到体积为j的背包中的最大总价值
          if(j >= vol)
          {
             dp[i][j] = maxs(dp[i][j],dp[i - 1][j - vol] + val[i]);
             //不要该物品以及要该物品两种情况
          }
       }
   }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        printf("%d\n",dp[n][v]);
    }
    return 0;
}

2、滚动数组

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn = 1005;
int n,v;
int vol;
int val[maxn];
int dp[maxn];
int maxs(int a,int b)
{
    return a > b ? a : b;
}
void init()
{
   memset(dp,0,sizeof(dp));
   scanf("%d %d",&n,&v);
   for(int i = 1;i <= n;i++)
   {
      scanf("%d",&val[i]);
   }
   for(int i = 1;i <= n;i++)
   {
       scanf("%d",&vol);
       for(int j = v;j >= 0;j--)
       {
           if(j >= vol)
           {
               dp[j] = maxs(dp[j],dp[j - vol] + val[i]);
               //滚动数组,dp[j]中相当于保存dp[i - 1][j],dp[j - vol]中相当于保存dp[i - 1][j - vol]
           }
       }
   }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        printf("%d\n",dp[v]);
    }
    return 0;
}