D - 英文题 （多组背包）

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.

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=4e4+;
typedef long long ll;
using namespace std;
struct node
{
    int h,a,c;
}p[maxn];
int dp[maxn];
bool cmp(node x,node y)
{
    return x.a<y.a;
}
int main()
{
    int K;
    cin>>K;
    for(int t=;t<=K;t++)
    {
       scanf("%d%d%d",&p[t].h,&p[t].a,&p[t].c);
    }
    sort(p+,p+K+,cmp);
    for(int t=;t<=K;t++)
    {
        for(int j=;j<=p[t].c;j++)
        {
            for(int k=p[t].a;k>=p[t].h;k--)
            {
                dp[k]=max(dp[k],dp[k-p[t].h]+p[t].h);
            }
        }
    }
    int ans=;
    for(int t=;t<=p[K].a;t++)
    {
        ans=max(ans,dp[t]);
    }
    printf("%d\n",ans);

    return ;
}