POJ2392：Space Elevator(多重背包)

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.

 

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了

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct node
{
    int h,hmax,n;
}a[500];

int cmp(node a,node b)
{
    return a.hmax<b.hmax;//按小到大排
}

int dp[400000],sum[400000];

int main()
{
    int t,i,j,k;
    while(~scanf("%d",&t))
    {
        for(i = 0;i<t;i++)
        scanf("%d%d%d",&a[i].h,&a[i].hmax,&a[i].n);
        sort(a,a+t,cmp);
        memset(dp,0,sizeof(dp));
        dp[0] = 1;
        int ans = 0;
        for(i = 0;i<t;i++)
        {
            memset(sum,0,sizeof(sum));
            for(j = a[i].h;j<=a[i].hmax;j++)
            {
                if(!dp[j] && dp[j-a[i].h] && sum[j-a[i].h]<a[i].n)
                {
                    dp[j] = 1;
                    sum[j] = sum[j-a[i].h]+1;//统计该石块放置的个数
                    if(ans<j)
                    ans = j;
                }
            }
        }
        printf("%d\n",ans);
    }

    return 0;
}