动态编程 - 油漆栅栏算法

Question

有一个有n个帖子的围栏,每个帖子都可以涂上一种k颜色.您必须绘制所有帖子,使得不超过两个相邻的栅栏柱具有相同的颜色.返回可以绘制栅栏的总方式.

diff - 具有不同颜色的组合的数量,
相同 - 具有相同颜色的组合的数量,
n - 发布的数量,
k - 颜色的数量.

对于n = 1:

diff = k;
same = 0;

对于n = 2:

diff = k * (k - 1);
same = k;

对于n = 3:

diff = (k + k * (k - 1)) * (k - 1);
same = k * (k - 1);

最后的公式是:

diff[i] = (diff[i - 1] + diff[i - 2]) * (k - 1);
same[i] =  diff[i - 1];

我理解如何找到same[i],但我不明白如何找到diff[i].你能解释一下这个公式diff[i]吗？

Answer 1

total[i] = diff[i] + same[i]   (definition)

diff[i] = (k - 1) * total[i-1]
        = (k - 1) * (diff[i-1] + same[i-1])
        = (k - 1) * (diff[i-1] + diff[i-2])

Answer 2

这是组合论的论点.

让我们根据问题陈述的规则diff[i, c]绘制i帖子的方式数量,以便最后一个围栏涂上颜色c.

我们有:

diff[i, c] = diff[i - 1, c'] + diff[i - 2, c''], c' != c OR c'' != c

对于c我们绘制的每一个i,前一个可以以a结束c' != c(在这种情况下我们不关心前一个是什么),或者第二个以前可以以a结束c'' != c(在这种情况下我们不关心以前是什么)是),或两者兼而有之.

有k - 1可能c' != c和k - 1为c'' != c.所以我们可以c从复发中删除并简单地写:

diff[i] = (k - 1) * (diff[i - 1] + diff[i - 2])

这是你拥有的.

Answer 3

该解决方案涉及详细的解释。请务必看一下。

public class PaintingFence {

  public static int paintFence(int n, int k) {
    //if there is 0 post then the ways to color it is 0.
    if(n == 0) return 0;

    //if there is one 1 post then the way to color it is k ways.
    if(n == 1) return k;

    /**
     * Consider the first two post.
     * case 1. When both post is of same color
     *    first post can be colored in k ways.
     *    second post has to be colored by same color.
     *    So the way in which the first post can be colored with same color is k * 1.
     *
     * case 2. When both post is of diff color
     *    first post can be colored in k ways.
     *    second post can be colored in k-1 ways.
     *    Hence the ways to color two post different is k * (k - 1)
     */
    int same = k * 1;
    int diff = k * (k -1);

    /**
     * As first 2 posts are already discussed, we will start with the third post.
     *
     * If the first two post are same then to make the third post different, we have
     * k-1 ways. Hence same * (k-1)
     * [same=k, so same * k-1 = k*(k-1) = diff => Remember this.]
     *
     * If the first two posts are different then to make the third different, we also have
     * k - 1 ways. Hence diff * (k-1)
     *
     * So to make third post different color, we have
     *  same * (k-1) + diff * (k-1)
     *  = (same + diff) * (k-1)
     *  = k-1 * (same + diff)
     */
    for(int i=3;i <=n; i++) {
      int prevDiff = diff;
      diff = (same + diff) * (k - 1); //as stated above

      /**
       * to make the third color same, we cannot do that because of constraint that only two
       * posts can be of same color. So in this case, we cannot have to same color so it has to be
       * diff.
       */
      same = prevDiff * 1;
    }

    return same + diff;
  }

  public static void main(String[] args) {
    System.out.println(paintFence(2, 4));
    System.out.println(paintFence(3, 2));
  }

}