pku 1185 炮兵阵地 压缩dp 解题报告

pku 1185 炮兵阵地 解题报告

思路:用二进制保存这一行中炮兵,例如1 0 0 0, 1表示炮兵.那么很明显用到了压缩dp。认真发现炮兵摆放的规律可知,每3行进行一个判断就可以了.也就是说,我们可以每3行都进行dp的转移,当然1、2行有特殊的处理.比如第i行为:1 0 0 0 0 第i+1行为0 0 0 0 1,那么怎么知道i+2的状态呢？其实我们可以先将合法的状态枚举出来,这样问题就变得简单,我们枚举i+2d状态时就与i行、i+1行的状态进行比较.看看这个是否合法,当然i行与i+1行合法,但不一定与i+2合法,这时我们就必须用dp[i][j][k]来进行每3行的dp了。假设当前行i,i-1行的状态为j,i+2行的状态为k.

算法:

1、  用map[i]保存i行存在山地的二进制状态;

2、  枚举一行（假设没有山地,也就是i行的枚举状态）的合法状态;

3、  特殊处理1与2行的情况;

4、  枚举判断当前行i的合法状态与map[i]的状态是否合法;再枚举i-1行,i-2行这3行之间的合法状态.

5、  合法的话,进行dp转移:

dp[i][j][k] = max(dp[i][j][k],dp[i - 1][k][0] + num[j]);

AC代码:

#include <stdio.h>

#include <string.h>

#define N 101

#define M 11

#define max(a, b) (a > b ? a : b)

 

//dp[i][j][k]表示在第i行的合法状态为j,第i-1行的合法状态为k时,炮兵的数量

int dp[N][61][61];

int num[N], map[N], state[N], n, m, count;

 

void init()

{

    int i, temp;

 

    count = 0; 

       //枚举一行中都可能出现的状态,从0到2^m - 1,判断其是否合法 

       //此函数的统计是假设当该行都为平地时,即共有cnt+1个合法状态

    for (i = 0; i < (1 << m); i++)      

    {

        temp = i;

        if (((temp << 1) & i) | ((temp << 2) & i))

              {

                     continue;                      //判断该行在这个状态时是否合法(任意炮兵都不在其他炮兵的攻击范围之内)

        }

              state[count] = i;                  //通过数组state[]记录合法的状态(十进制表示)

        num[count] = temp & 1;             //num[]数组记录这个合法状态下'1'的个数(也就是炮兵的个数)

        while (temp = (temp >> 1))

              {

                     num[count] += temp & 1;

              }

        count++;                         

    }   

}

 

void solve()

{

       int i, j, k, p;

 

       for (i = 0; i < n; i++)

       {

              for (j = 0; j < count; j++)

              {

                     //由于state[]记录的是一行的合法状态,那么用map[]&state[],则可以判断map[i]与state[i]是合法的

                     if (map[i] & state[j])

                     {

                            continue;

                     }

                     //i=0与i=1为特殊处理

                     if (i == 0)

                     {

                            dp[i][j][0] = num[j];

                     }

                     else if (i == 1)

                     { 

                            for (k = 0; k < count; k++)                  //枚举第0行的可能状态

                {

                                   if (state[j] & state[k])

                                   {

                                          continue;                            //判断上下两行(0和1)的合法状态是否兼容彼此

                    }

                                   dp[i][j][k] = max(dp[i][j][k], dp[i-1][k][0]+num[j]);

                }

                     }

                     else

                     {

                            //枚举第i行的所有可能状态

                            for (k = 0; k < count; k++)

                            {

                                   //判断上下两行(i和i-1)的合法状态是否兼容彼此

                                   if (state[j] & state[k])

                                   {

                                          continue;

                                   }

                                   for (p = 0; p < count; p++)

                                   {

                                          //判断分别i行与i-2行的合法状态、i-1行与i-2行都是否彼此兼容

                                          if (state[k] & state[p] || state[j] & state[p])

                                          {

                                                 continue;

                                          }

                                          dp[i][j][k] = max(dp[i][j][k], dp[i-1][k][p]+num[j]);

                                   }

                            }

                     }

              }

       }

       int ans = 0;

       for (j = 0; j < count; j++)

       {

              for (k = 0; k < count; k++)

              {

                     if (dp[n - 1][j][k] > ans)

                     {

                            ans = dp[n - 1][j][k];

                     }

              }

       }

       printf("%d/n", ans);      

}

 

int main()

{

       freopen("1.txt", "r", stdin);

       int i, j;

       char s[11];

 

       scanf("%d%d", &n, &m);

       for (i = 0; i < n; i++)

       {

              scanf("%s", s);

              //将每一行出现的山地存储在二进制中

              for (j = 0; j < m; j++)

              {

                     if (s[j] == 'H')

                     {

                            map[i] += (1 << (m - 1 - j));

                     }

              }

       }

       init();

       solve();

 

       return 0;

}