ACM: 动态规划题 poj&nb…

Space
Elevator
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

 

题意: cows们打算去太空, 它们有K种不同types的材料建造升降舱, 每种材料有c_i个和

      每个高度h_i,
但是有一个限制: 当用这种材料结束时升降舱高度不超过a_i. 现在

     
要你求出升降舱最大高度.

 

解题思路:

     
1. 明显背包问题, 并且是多重背包. 设状态dp[i][j]: 前面i中材料是否能不超过限制条件

        
下(j背包容量), 放入背包中.

     
2. 问题跟a_i有大小有关, 应该先对a_i排序, 小的先用.

     
3. 因为每次使用新的材料只跟上一种有关, dp[i][j]降至一维dp[j]即可.

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

#define MAX 40005

#define MAXSIZE 405

struct node

{

 int h, al, c;

}block[MAXSIZE];

int n;

bool dp[MAX];

inline int max(int a, int b)

{

 return a > b ? a : b;

}

bool cmp(node &a, node &b)

{

 return a.al < b.al;

}

int DP()

{

 int result = 0;

 dp[0] = true;

 for(int i = 1; i <= n; ++i)

 {

  for(int j = 1; j
<= block[i].c; ++j)

  {

   for(int k =
result; k >= 0; --k)

   {

    if(
dp[k] )

    {

     if(
k+block[i].h <= block[i].al )

     {

      dp[
k+block[i].h ] = true;

      result
= max(result, k+block[i].h);

     }

    }

   }

  }

 }

 return result;

}

int main()

{

// freopen("input.txt", "r", stdin);

 while(scanf("%d", &n) !=
EOF)

 {

  for(int i = 1; i
<= n; ++i)

   scanf("%d %d
%d", &block[i].h, &block[i].al,
&block[i].c);

  sort(block+1, block+n+1,
cmp);

  memset(dp, false,
sizeof(dp));

  printf("%d\n", DP());

 }

 return 0;

}