题目：
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：一群牛要修建高楼，他们手头有n种建筑材料，之后交代建筑材料的高度，能达到的最大高度（超过就会损坏），数量，求解高楼的最大高度。
解题思路：多重背包，先对建筑材料的能承受的最大高度排序，将承受最大高度小的材料优先放置，才能使结构达到最优，高度达到最大，之后就是多重背包的模板求解。
ac代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define maxn 100005
using namespace std;
int cnt[maxn];
int dp[maxn];
struct node{
int ht;
int num;
int maxx;
}co[maxn];
bool cmp(node a,node b)
{
return a.maxx<b.maxx;
}
int main()
{
int k;
while(scanf("%d",&k)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=k;i++)
{
scanf("%d%d%d",&co[i].ht,&co[i].maxx,&co[i].num);
}
sort(co+1,co+1+k,cmp);
dp[0]=1;
int ans=0;
for(int i=1;i<=k;i++)
{
memset(cnt,0,sizeof(cnt));
for(int j=co[i].ht;j<=co[i].maxx;j++)
{
if(!dp[j]&&dp[j-co[i].ht]&&cnt[j-co[i].ht]+1<=co[i].num)
{
dp[j]=1;
cnt[j]=cnt[j-co[i].ht]+1;
if(ans<j) ans=j;
}
}
}
printf("%d\n",ans);
}
return 0;
}