Space Elevator （多重背包问题）

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.

PS：看题的时候不知道怎么思考，但是，将所有的块数按最大的限制长度重小到大排序后，就会发现这是一个多重背包问题，但是也可以用完全背包做，而且完全背包的时间复杂度还小一些，下面给出两种做法的代码。

多重背包

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
const int maxn= 5e2+5;
const int mod=1e9;
#define me(a) memset(a,0,sizeof(a))
#define ll long long
using namespace std;
struct node
{
    int h,m,c;
    bool friend operator<(node a,node b)
    {
        return a.m<b.m;
    }
};
int main()
{
    node a[maxn];
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        scanf("%d%d%d",&a[i].h,&a[i].m,&a[i].c);
    sort(a,a+n);
    int dp[40005],num[40005];me(dp),me(num);
    int ma=0; 
    for(int i=0;i<n;i++)
    {
      for(int t=0;t<a[i].c;t++)
        for(int j=a[i].m;j>=a[i].h;j--)
        {
            dp[j]=max(dp[j-a[i].h]+a[i].h,dp[j]);
            if(dp[j]>ma)
            	ma=dp[j];
        }
    }
        cout<<ma<<endl;
    return 0;
}

完全背包

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<string>
#include<map>
const int maxn= 5e2+5;
const int mod=1e9;
#define me(a) memset(a,0,sizeof(a))
#define ll long long
using namespace std;
struct node
{
    int h,m,c;
    bool friend operator<(node a,node b)
    {
        return a.m<b.m;
    }
};
int main()
{
    node a[maxn];
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        scanf("%d%d%d",&a[i].h,&a[i].m,&a[i].c);
    sort(a,a+n);
    int dp[40005],num[40005];me(dp);
    int maxn=0;
    dp[0]=1;
    for(int i=0;i<n;i++)
    {
        me(num);
        for(int j=a[i].h;j<=a[i].m;j++)
        {
            if(dp[j-a[i].h]&&!dp[j]&&num[j-a[i].h]+1<=a[i].c)
                {
                    dp[j]=1;
                    num[j]=num[j-a[i].h]+1;
                   if(j>maxn)
                        maxn=j;
                }
        }
    }
        cout<<maxn<<endl;
    return 0;
}