ACM: 动态规划题 poj 2923

[align=center]Relocation[/align]
Description

Emma and Eric are moving to their new house they bought after
returning from their honeymoon. Fortunately, they have a few
friends helping them relocate. To move the furniture, they only
have two compact cars, which complicates everything a bit. Since
the furniture does not fit into the cars, Eric wants to put them on
top of the cars. However, both cars only support a certain weight
on their roof, so they will have to do several trips to transport
everything. The schedule for the move is planed like this:

At their old place, they will put furniture on both cars.

Then, they will drive to their new place with the two cars and
carry the furniture upstairs.

Finally, everybody will return to their old place and the
process continues until everything is moved to the new place.

Note, that the group is always staying together so that they can
have more fun and nobody feels lonely. Since the distance between
the houses is quite large, Eric wants to make as few trips as
possible.

Given the weights wi of each individual piece
of furniture and the capacities C1 and
C2 of the two cars, how many trips to the new
house does the party have to make to move all the furniture? If a
car has capacity C, the sum of the weights of all the
furniture it loads for one trip can be at most C.

Input

The first line contains the
number of scenarios. Each scenario consists of one line containing
three numbers n, C1 and
C2. C1 and C2
are the capacities of the cars (1 ≤ Ci ≤ 100) and
n is the number of pieces of furniture (1 ≤ n ≤ 10).
The following line will contain n integers
w1, …, wn, the weights of the
furniture (1 ≤ wi ≤ 100). It is guaranteed that
each piece of furniture can be loaded by at least one of the two
cars.

Output

The output for every
scenario begins with a line containing “Scenario
#i:”, where i is the number of the
scenario starting at 1. Then print a single line with the number of
trips to the new house they have to make to move all the furniture.
Terminate each scenario with a blank line.

Sample Input

2

6 12 13

3 9 13 3 10 11

7 1 100

1 2 33 50 50 67 98

Sample Output

Scenario #1:

2

Scenario #2:

3

 

题意: Emma and Eric 蜜月回来, 他们有2部车载重量分别为c1,c2.

      有n件家俱,
每件家俱种类为w[i], 现在要用最少的次数运完全部的家俱.

 

解题思路: (这题没有思路, 从网上学会的)

    
1. 物品种数是n个, 选取的状态有2^n种选择方式,
(1<=n<=10)范围不大.

       
先将可以用2车运送的选择方式求出来, 这是一次背包.

       
分析:

       
(1). 状态用二进制表示, 压缩状态嘛, 每次枚举一个状态, 0表示不选取,

            
1表示选取. 显然, 背包关键值是c1或则c2, 每次的区间是[c1, w[i]]

            
或则[c2, w[i]];

        (2).
最后是判断这次状态枚举是否可行.

    
2. 在第一次背包保存下来的可行状态中, 接下来就是计算最少运算次数.

       
设状态dp[i]表示i是二进制, 0表示未运, 1表示已经运, 在i状态下最少的运送次数.

        当前状态j不与状态situation[i]没有重复(因为每件物品只运送一次)时:

       
dp[ j|situation[i] ] = min(dp[
j|situation[i] ], dp[j]+1);

       
(j|situation[i]表示两个状态的合并)

 

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 11

const int INF = (1<<29);

int n, c1, c2;

int w[MAX], s;

int dp[1<<MAX], num,
flag[1<<MAX];

inline int min(int a, int b)

{

 return a < b ?  a
: b;

}

bool judge(int x)

{

 bool
temp[1<<MAX];

 int i, j, sum = 0;

 memset(temp, false, sizeof(temp));

 temp[0] = 1;

 for(i = 0; i < n; ++i)

 {

  if( x &
(1<<i) )

  {

   sum +=
w[i];

   for(j = c1; j
>= w[i]; --j)

   {

    if(
temp[ j-w[i] ] )

     temp[j]
= true;

   }

  }

 }

 for(i = 0; i <= c1; ++i)

 {

  if( temp[i]
&& sum-i <= c2
)

   return
true;

 }

 return false;

}

int DP()

{

 dp[0] = 0;

 for(int i = 0; i < num; ++i)

 {

  for(int j = s-1-flag[i]; j
>= 0; --j)

  {

   if(
!(j&flag[i]) ) //每个物品只运送一次.

   {

    dp[
j|flag[i] ] = min( dp[ j|flag[i] ], dp[j]+1 );

   }

  }

 }

 return dp[s-1];

}

int main()

{

// freopen("input.txt", "r", stdin);

 int i, caseNum, Num = 1;

 scanf("%d", &caseNum);

 while(caseNum--)

 {

  num = 0;

  scanf("%d %d %d",
&n, &c1,
&c2);

  for(i = 0; i <
n; ++i)

   scanf("%d",
&w[i]);

  

  s =
1<<n;

  for(i = 0; i <
s; ++i)

  {

   if( judge(i)
)

    flag[
num++ ] = i;

   dp[i] =
INF;

  }

  printf("Scenario
#%d:\n%d\n\n", Num++, DP());

 }

 return 0;

}