https://cn.vjudge.net/problem/POJ-2392

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：有n种石头，每个石头的高度hi，允许用的最高高度为ai，数量为ci，求能组合的最高高度。

 

思路：先给这些石头排个序，最高高度小的在前面，这样才能使得高度最大，然后再按多重背包来做，这样就可以了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#include<vector>
#include<set>
#include<list>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
struct node{
	int h,a,c;
}p[405];
bool cmp(node a,node b){
	return a.a<b.a;
}
int h[4005],c[4005],a[4005];
int dp[40005];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
    	scanf("%d%d%d",&p[i].h,&p[i].a,&p[i].c);
	}
	sort(p+1,p+1+n,cmp);
	int cnt=1;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=p[i].c;j<<=1){
			h[cnt]=j*p[i].h;
			c[cnt]=1;
			a[cnt]=p[i].a;
			cnt++;
			p[i].c-=j;
		}
		if(p[i].c){
			h[cnt]=p[i].c*p[i].h;
			c[cnt]=1;
			a[cnt]=p[i].a;
			cnt++;
		}
	}
	for(int i=1;i<cnt;i++){
		for(int j=a[i];j>=h[i];j--){
			dp[j]=max(dp[j],dp[j-h[i]]+h[i]);
		}
	}
	for(int i=40000;i>=0;i--){
		if(dp[i]==i){
			printf("%d\n",i);
			break;
		}
	}
	return 0;
}