动态规划——油漆栅栏算法

Question

有一个nposts的围栏，每个post可以用k[=40]中的一种来画=] 颜色。您必须绘制所有 post，这样相邻的栅栏不超过两个 post 具有相同的颜色。 Return 绘制围栏的方式总数。

diff - 不同颜色的组合数，
相同 - 相同颜色的组合数，
n - post 的数量，
k - 颜色数。

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];

我知道如何找到 same[i]，但我不明白如何找到 diff[i]。你能解释一下 diff[i] 的公式吗？

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

这是一个组合论证。

设 diff[i, c] 为根据问题陈述的规则绘制 i 个帖子的方法数，使得最后一个栅栏被涂上颜色 c.

我们有：

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

对于我们绘制 i 的每个 c，前一个可以以 c' != c 结尾（在这种情况下我们不关心第二个前一个是什么），或者前一个可以以 c'' != c 结尾（在这种情况下我们不关心前一个是什么），或者两者兼而有之。

c' != c 有 k - 1 种可能性，c'' != c 有 k - 1 种可能性。所以我们可以从循环中删除 c 并简单地写：

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

你有什么。

Answer 3

解决方案有详细的解释。请一定要看看

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));
  }

}

动态规划——油漆栅栏算法

Dynamic programming - paint fence algorithm

algorithm

dynamic-programming

combinatorics