POJ 2392 Space Elevator

朱雀 2022-06-18 07:25 192阅读 0赞

Space Elevator  
Time Limit: 1000MS Memory Limit: 65536K  
Total Submissions: 11622 Accepted: 5518  
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

48  
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

> 题意：一群牛要上太空，给出n种石块，每种石块给出单块高度，总高度不能超过的最大值，数量，要求用这些石块能组成的最大高度  
> 思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包

> 思路：在进行多重背包之前要进行一次排序，将最大高度小的放在前面，只有这样才能得到最优解，如果将大的放在前面，后面有的小的就不能取到，排序之后就可以进行完全背包

#include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    
    using namespace std;
    const int maxn = 400000;
    struct node
    {
        int h_i;
        int a_i;
        int c_i;
        bool operator < (const node &n)const
        {
            return a_i < n.a_i;
        }
    }vis[405];
    
    int dp[maxn];
    int cnt[maxn];
    int main()
    {
        int k;
        while(cin>>k)
        {
            for(int i=0;i<k;i++)
            {
                cin>>vis[i].h_i>>vis[i].a_i>>vis[i].c_i;
            }
            sort(vis,vis+k);
            memset(dp,0,sizeof(dp));
            dp[0] = 1;
            int ans = 0;
            for(int i=0;i<k;i++)
            {
                memset(cnt,0,sizeof(cnt));
                for(int j=vis[i].h_i;j<=vis[i].a_i;j++)
                {
                    // 维持单调队列的递减特性
                    if(!dp[j]&&dp[j-vis[i].h_i]&&cnt[j-vis[i].h_i]<vis[i].c_i)
                    {
                        dp[j]=1;
                        cnt[j]=cnt[j-vis[i].h_i]+1;
                        if(ans<j)
                            ans=j;
                    }
                }
            }
            cout<<ans<<endl;
        }
        return 0;
    }