Problem 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
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题意：给出 k 块砖，以及每块砖的高度 h，数量 c 以及第 i 块砖不能超过高度 a 垒放，问最高可以垒多高。
思路：先将砖头按照不能超过的高度排序，随后套用多重背包模版即可。
Source Program
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 100001
#define MOD 123
#define E 1e-6
using namespace std;
int dp[N];
int num[N];
struct Node{
int h;
int c;
int a;
}block[N];
int cmp(Node x,Node y)
{
return x.a<y.a;
}
int main()
{
int k;
scanf("%d",&k);
int height=0;
for(int i=0;i<k;i++)
{
scanf("%d%d%d",&block[i].h,&block[i].a,&block[i].c);
height=max(block[i].a,height);
}
sort(block,block+k,cmp);
for(int i=0;i<N;i++)
dp[i]=-INF;
//memset(dp,-INF,sizeof(dp));
dp[0]=0;
for(int i=0;i<k;i++)
{
for(int j=0;j<=height;j++)
num[j]=0;
//memset(num,0,sizeof(num));
for(int j=block[i].h;j<=block[i].a;j++)
{
if(dp[j]==-INF&&dp[j-block[i].h]>-INF)
{
if(num[j-block[i].h]<block[i].c)
{
dp[j]=j;
num[j]=num[j-block[i].h]+1;
}
}
}
}
int maxx=0;
for(int i=height;i>=0;i--)
{
if(dp[i]>-INF)
{
maxx=i;
break;
}
}
printf("%d\n",maxx);
return 0;
}
