Space Elevator

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 12308	Accepted: 5854

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

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.

Source

題意: 一群牛想要上太空，有一些磚塊，用來搭建上太空的路，因為宇宙射線線的存在，每塊磚都有其最高限度，不能超過其最高限度。

           給你這些磚塊的高度， 最大限度，數量。

思路：Dp + 貪心,  先按照最大限度排序（升序），然後列舉每一類磚塊，在已存在高度繼續往上壘的情況，不斷記錄下下來。

程式碼：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef struct block{
	int h, a, c;// 高度 ，限度，數量 
}B;
B a[410];
bool dp[40100];
// dp[i] == true 高度 i 存在 
bool cmp(B m,B n)
{
	return m.a < n.a;
}
int main()
{
	int k;
	scanf("%d", &k);
	int max = 0;
	for(int i=1; i<=k; i++)
	{
		scanf("%d %d %d",&a[i].h, &a[i].a, &a[i].c);
	}
	
	sort(a+1, a+k+1, cmp); // 按照限度排序 
	memset(dp, 0, sizeof(dp));
	dp[0] = 1;
	
	for(int i=1; i<=k; i++)
	{
	     for(int j=a[i].a; j>=0; j--) 
		 // 從每一類磚的限度往下列舉 ，如果從0往上需要 01 陣列 
	     { 	
	     	 if(dp[j])  
	     	 {
                  for(int l=1; l<=a[i].c; l++)
                  {
                  	   int tem = j + a[i].h * l;
                  	   if(tem <= a[i].a)
                  	   {
                  	        dp[tem] = 1;	
					   }
				  }
             }
		 }
    }
    
    int ans = 0;
    for(int i=0; i<=a[k].a; i++)
    {
    	if(dp[i])  ans = i;
	}
    printf("%d\n", ans);
	return 0;
}