POJ 2392-Space Elevator(多重部分和-多重背包)

雨点打透心脏的1/2处 2022-07-24 11:19 97阅读 0赞

# Space Elevator #

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 10506</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 4994</td> 
  </tr> 
 </tbody> 
</table>

Description

The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has height h\_i (1 <= h\_i <= 100) and is available in quantity c\_i (1 <= c\_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type i can exceed a maximum altitude a\_i (1 <= a\_i <= 40000).  
  
Help the cows build the tallest space elevator possible by stacking blocks on top of each other according to the rules.

Input

\* Line 1: A single integer, K  
  
\* Lines 2..K+1: Each line contains three space-separated integers: h\_i, a\_i, and c\_i. Line i+1 describes block type i.

Output

\* Line 1: A single integer H, the maximum height of a tower that can be built

Sample Input

3
    7 40 3
    5 23 8
    2 52 6

Sample Output

Hint

OUTPUT DETAILS:  
  
From the bottom: 3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 is not legal, since the top of the last type 1 block would exceed height 40.

Source

[USACO 2005 March Gold][]

## 题目意思： ##

有K组数，后面三个数h，a，c分别表示一块砖的高度、用这种砖的最大高度和用这种砖的最多块数。  
求用这些砖叠在一起的最大高度。

## 解题思路： ##

用结构体根据“用这种砖的最大高度”升序排列，然后有两种思路：  
①多重部分和；  
②多重背包，限制高度作为背包中的最大高度。  
  
两种方法解题占用内存一样，但是①比②快一倍。  
  
AC代码①多重部分和

#include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define MAXN 100010
    
    int dp[MAXN];//d[i]表示石块能组成的高度为i
    
    struct Node
    {
        int h,a,c;//分别表示一块砖的高度、用这种砖的最大高度和最多块数
    } s[MAXN];
    int cmp(Node x,Node y)//结构体排序
    {
        return x.a<y.a;//升序
    }
    int main()
    {
        int n,i,j,ans=0;
        cin>>n;
        for(i=0; i<n; ++i)
            cin>>s[i].h>>s[i].a>>s[i].c;
        sort(s,s+n,cmp);//按最大高度升序排列
        memset(dp,-1,sizeof dp);
        dp[0]=0;
        for(i=0; i<n; i++)//多重部分和
            for(j=0; j<=s[i].a; j++)
            {
                if (dp[j]>=0) dp[j]=s[i].c;
                else if (j<s[i].h || dp[j-s[i].h] <=0) dp[j]=-1;
                else dp[j] = dp[j-s[i].h]-1;
            }
        for (i=s[n-1].a; i>=0; --i)
            if(dp[i]>=0)
            {
                ans=i;
                break;
            }
        cout<<ans<<endl;
        return 0;
    }

AC代码②多重背包

#include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define MAXN 100010//一开始开了10010一直RE…囧o(╯□╰)o
    
    int dp[MAXN];//d[i]表示石块能组成的高度为i
    
    struct Node
    {
        int h,a,c;//分别表示一块砖的高度、用这种砖的最大高度和最多块数
    } s[MAXN];
    int cmp(Node x,Node y)//结构体排序
    {
        return x.a<y.a;//升序
    }
    int main()
    {
        int n,i,j,k,ans=0;
        cin>>n;
        for(i=0; i<n; ++i)
            cin>>s[i].h>>s[i].a>>s[i].c;
        sort(s,s+n,cmp);//按最大高度升序排列
        memset(dp,0,sizeof(dp));//初始化
        for(i=0; i<n; ++i)//n种砖块，使用多重背包
        {
            if(s[i].h*s[i].c>=s[i].a)//完全背包
                for(j=s[i].h; j<=s[i].a; j++)
                    dp[j]=max(dp[j],dp[j-s[i].h]+s[i].h);
            else//01背包
            {
                for(k=1; k<=s[i].c; k<<=1) //数量
                {
                    for(j=s[i].a; j>=k*s[i].h; --j)//高度
                        dp[j]=max(dp[j],dp[j-k*s[i].h]+k*s[i].h);
                    s[i].c-=k;//减去用过的数量
                }
                for(j=s[i].a; j>=s[i].h*s[i].c; --j)
                    dp[j]=max(dp[j],dp[j-s[i].h*s[i].c]+s[i].h*s[i].c);
            }
        }
        for(i=1; i<=s[n-1].a; ++i)//s[n-1].a表示所有限制高度中的最大值
            ans=max(ans,dp[i]);//找出石块能组成的最大高度
        cout<<ans<<endl;
        return 0;
    }

[USACO 2005 March Gold]: http://poj.org/searchproblem?field=source&key=USACO+2005+March+Gold