给定两个正整数N和C。有一个2 * N矩阵，其中矩阵的每个单元格都可以用0或1进行着色。任务是寻找多种方式，形成的2 * N矩阵恰好具有c个成分。
A cell is said to be in the same component with the other cell if it shares a side with the other cell (immediate neighbor) and is of the same color.
例子：
Input: N = 2, C = 1
Output: 2
Explanation:
C = 1 can be possible when all the cells of the matrix are colored in the same color.
We have 2 choices for coloring 0 and 1. Hence the answer is 2.
Input: N = 1, C = 2
Output: 2
方法：解决此问题的想法是使用动态编程。构造一个4D DP矩阵
大小，其中N是列数，C是组件数，2 * 2维是最后一列的排列。最后一列可以有四个不同的组合。它们是00、01、10、11。
当最后一列的排列为row1，row2时，矩阵dp [i] [j] [row1] [row2]的每个单元给出在i列中具有j个分量的方式的数量。如果已经评估了当前子问题，即dp [column] [component] [row1] [row2]！= -1，则使用此结果，否则递归计算该值。
- 基本情况：当列数大于N时，返回0。如果列数等于N，则答案取决于组件数。如果分量等于C，则返回1，否则返回0。
- 情况1 ：当前一列的行中具有相同的颜色({0，0}或{1，1})时。
在这种情况下，如果当前列在行中具有不同的值({0，1}或{1，0})，则组件将增加1。如果当前列的颜色相反，但行中的值相同，则分量也将增加1。也就是说，当前一列具有{0，0}时，当前列具有{1，1}。或当前列为{1，1}时当前列为{0，0}。当当前列与上一列具有相同的组合时，组件将保持不变。 - 情况2 ：当前一列的行中具有不同的颜色({0，1}或{1，0})时。
在这种情况下，如果当前列与上一列具有完全相反的组合，则组件将增加2。那就是当当前列是{1，0}而前一列是{0，1}时。或当前列为{0，1}，前一列为{1，0}。在所有其他情况下，组件保持不变。
下面是上述方法的实现：
C++
// C++ implementation to find the
// number of ways to make exactly
// C components in a 2 * N matrix
#include
using namespace std;
// row1 and row2 are one
// when both are same colored
int n, k;
int dp[1024][2048][2][2];
// Function to find the number of
// ways to make exactly C components
// in a 2 * N matrix
int Ways(int col, int comp,
int row1, int row2)
{
// if No of components
// at any stage exceeds
// the given number
// then base case
if (comp > k)
return 0;
if (col > n) {
if (comp == k)
return 1;
else
return 0;
}
// Condition to check
// if already visited
if (dp[col][comp][row1][row2] != -1)
return dp[col][comp][row1][row2];
// if not visited previously
else {
int ans = 0;
// At the first column
if (col == 1) {
// color {white, white} or
// {black, black}
ans
= (ans
+ Ways(col + 1, comp + 1, 0, 0)
+ Ways(col + 1, comp + 1, 1, 1));
// Color {white, black} or
// {black, white}
ans
= (ans
+ Ways(col + 1, comp + 2, 0, 1)
+ Ways(col + 1, comp + 2, 1, 0));
}
else {
// If previous both
// rows have same color
if ((row1 && row2)
|| (!row1 && !row2)) {
// Fill with {same, same} and
// {white, black} and {black, white}
ans = (((ans
+ Ways(col + 1, comp + 1, 0, 0))
+ Ways(col + 1, comp + 1, 1, 0))
+ Ways(col + 1, comp + 1, 0, 1));
// Fill with same without
// increase in component
// as it has been
// counted previously
ans = (ans
+ Ways(col + 1, comp, 1, 1));
}
// When previous rows
// had {white, black}
if (row1 && !row2) {
ans = (((ans
+ Ways(col + 1, comp, 0, 0))
+ Ways(col + 1, comp, 1, 1))
+ Ways(col + 1, comp, 1, 0));
ans = (ans
+ Ways(col + 1, comp + 2, 0, 1));
}
// When previous rows
// had {black, white}
if (!row1 && row2) {
ans = (((ans
+ Ways(col + 1, comp, 0, 0))
+ Ways(col + 1, comp, 1, 1))
+ Ways(col + 1, comp, 0, 1));
ans = (ans
+ Ways(col + 1, comp + 2, 1, 0));
}
}
// Memoization
return dp[col][comp][row1][row2] = ans;
}
}
// Driver Code
signed main()
{
n = 2;
k = 1;
memset(dp, -1, sizeof(dp));
// Initially at first coloumn
// with 0 components
cout << Ways(1, 0, 0, 0);
return 0;
} Java
// Java implementation to find the
// number of ways to make exactly
// C components in a 2 * N matrix
class GFG{
// row1 and row2 are one
// when both are same colored
static int n, k;
static int [][][][]dp = new int[1024][2048][2][2];
// Function to find the number of
// ways to make exactly C components
// in a 2 * N matrix
static int Ways(int col, int comp,
int row1, int row2)
{
// if No of components
// at any stage exceeds
// the given number
// then base case
if (comp > k)
return 0;
if (col > n)
{
if (comp == k)
return 1;
else
return 0;
}
// Condition to check
// if already visited
if (dp[col][comp][row1][row2] != -1)
return dp[col][comp][row1][row2];
// if not visited previously
else
{
int ans = 0;
// At the first column
if (col == 1)
{
// color {white, white} or
// {black, black}
ans = (ans + Ways(col + 1, comp + 1, 0, 0) +
Ways(col + 1, comp + 1, 1, 1));
// Color {white, black} or
// {black, white}
ans = (ans + Ways(col + 1, comp + 2, 0, 1) +
Ways(col + 1, comp + 2, 1, 0));
}
else
{
// If previous both
// rows have same color
if ((row1 > 0 && row2 > 0) ||
(row1 == 0 && row2 ==0))
{
// Fill with {same, same} and
// {white, black} and {black, white}
ans = (((ans +
Ways(col + 1, comp + 1, 0, 0)) +
Ways(col + 1, comp + 1, 1, 0)) +
Ways(col + 1, comp + 1, 0, 1));
// Fill with same without
// increase in component
// as it has been
// counted previously
ans = (ans +
Ways(col + 1, comp, 1, 1));
}
// When previous rows
// had {white, black}
if (row1 > 0 && row2 == 0)
{
ans = (((ans +
Ways(col + 1, comp, 0, 0)) +
Ways(col + 1, comp, 1, 1)) +
Ways(col + 1, comp, 1, 0));
ans = (ans +
Ways(col + 1, comp + 2, 0, 1));
}
// When previous rows
// had {black, white}
if (row1 ==0 && row2 > 0)
{
ans = (((ans +
Ways(col + 1, comp, 0, 0)) +
Ways(col + 1, comp, 1, 1)) +
Ways(col + 1, comp, 0, 1));
ans = (ans +
Ways(col + 1, comp + 2, 1, 0));
}
}
// Memoization
return dp[col][comp][row1][row2] = ans;
}
}
// Driver Code
public static void main(String[] args)
{
n = 2;
k = 1;
for (int i = 0; i < 1024; i++)
for (int j = 0; j < 2048; j++)
for (int k = 0; k < 2; k++)
for (int l = 0; l < 2; l++)
dp[i][j][k][l] = -1;
// Initially at first coloumn
// with 0 components
System.out.print(Ways(1, 0, 0, 0));
}
}
// This code is contributed by Rajput-JiC#
// C# implementation to find the
// number of ways to make exactly
// C components in a 2 * N matrix
using System;
class GFG{
// row1 and row2 are one
// when both are same colored
static int n, k;
static int [,,,]dp = new int[ 1024, 2048, 2, 2 ];
// Function to find the number of
// ways to make exactly C components
// in a 2 * N matrix
static int Ways(int col, int comp,
int row1, int row2)
{
// If No of components
// at any stage exceeds
// the given number
// then base case
if (comp > k)
return 0;
if (col > n)
{
if (comp == k)
return 1;
else
return 0;
}
// Condition to check
// if already visited
if (dp[ col, comp, row1, row2 ] != -1)
return dp[ col, comp, row1, row2 ];
// If not visited previously
else
{
int ans = 0;
// At the first column
if (col == 1)
{
// color {white, white} or
// {black, black}
ans = (ans + Ways(col + 1, comp + 1, 0, 0) +
Ways(col + 1, comp + 1, 1, 1));
// Color {white, black} or
// {black, white}
ans = (ans + Ways(col + 1, comp + 2, 0, 1) +
Ways(col + 1, comp + 2, 1, 0));
}
else
{
// If previous both
// rows have same color
if ((row1 > 0 && row2 > 0) ||
(row1 == 0 && row2 == 0))
{
// Fill with {same, same} and
// {white, black} and {black, white}
ans = (((ans +
Ways(col + 1, comp + 1, 0, 0)) +
Ways(col + 1, comp + 1, 1, 0)) +
Ways(col + 1, comp + 1, 0, 1));
// Fill with same without
// increase in component
// as it has been
// counted previously
ans = (ans +
Ways(col + 1, comp, 1, 1));
}
// When previous rows
// had {white, black}
if (row1 > 0 && row2 == 0)
{
ans = (((ans +
Ways(col + 1, comp, 0, 0)) +
Ways(col + 1, comp, 1, 1)) +
Ways(col + 1, comp, 1, 0));
ans = (ans +
Ways(col + 1, comp + 2, 0, 1));
}
// When previous rows
// had {black, white}
if (row1 == 0 && row2 > 0)
{
ans = (((ans +
Ways(col + 1, comp, 0, 0)) +
Ways(col + 1, comp, 1, 1)) +
Ways(col + 1, comp, 0, 1));
ans = (ans +
Ways(col + 1, comp + 2, 1, 0));
}
}
// Memoization
return dp[ col, comp, row1, row2 ] = ans;
}
}
// Driver Code
public static void Main(String[] args)
{
n = 2;
k = 1;
for(int i = 0; i < 1024; i++)
for(int j = 0; j < 2048; j++)
for(int K = 0; K < 2; K++)
for(int l = 0; l < 2; l++)
dp[ i, j, K, l ] = -1;
// Initially at first coloumn
// with 0 components
Console.Write(Ways(1, 0, 0, 0));
}
}
// This code is contributed by Rajput-Ji输出：
2
时间兼容性： O(N * C)