[Codeforces] 1051D Bicolorings

(color{red}{mathcal{Description}})

You are given a grid, consisting of (2) rows and (n) columns. Each cell of this grid should be colored either black or white.

Two cells are considered neighbours if they have a common border and share the same color. Two cells (A) and (B) belong to the same component if they are neighbours, or if there is a neighbour of (A) that belongs to the same component with (B) .

Let's call some bicoloring beautiful if it has exactly (k) components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo (998244353) .

给定一个(2 imes n) 的棋盘，可以对上面的格子黑白染色，求染色后棋盘上的联通块的个数正好为 (k) 的染色方案数

(color{red}{mathcal{Input Format}})

The only line contains two integers (n) and (k) — the number of columns in a grid and the number of components required.

一行两个整数 (n) , (k)

(color{red}{mathcal{Output Format}})

Print a single integer — the number of beautiful bicolorings modulo (998244353) .

一个整数，表示方案数(mod 998244353) 后的结果

(color{red}{mathcal{DataSize Agreement}})

$1≤n≤1000 $, (1 le k le 2n)

(color{red}{mathcal{Solution}})

题目简洁明了（大意），考虑用动态规划

我们不妨先从此题的弱化版想起，也就是只有一行的时候

令 (dp[i][j][k]) 表示到第 (i) 列，产生了 (j) 个连通块，状态为 (k) 的时候的方案数，显然 (k) 只有两种状态: (0) 或 (1)，我们可以得到转移方程（最好画图理解）:

[dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j-1][1] ]

[dp[i][j][1]=dp[i-1][j-1][0]+dp[i-1][j][1] ]

理解了上面这个转移方程，对于此题只是变成了 (4) 个状态，转移方程也自然出来了（最好画图理解）

[k:00, 10, 01, 11 ]

[dp[i][j][0]=dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j-1][3] ]

[dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j-2][2]+dp[i-1][j-1][3]) ]

[dp[i][j][2]=(dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][2]+dp[i-1][j-1][3]) ]

[dp[i][j][3]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j][3]) ]

[2leq i leq N, 1 leq j leq 2i ]

初始化 (dp[1][1][0]=dp[1][2][1]=dp[1][2][2]=dp[1][1][3] = 1)

(color{red}{mathcal{Code}})

#include <bits/stdc++.h>
#define LL long long
#define reg register

using namespace std;

const int kM = 1010, kC = 2010, kS = (1 << 1) + 2, mod = 998244353;

LL dp[kM][kC][kS] = {0}; //kM-列 kC-个数 kS-00 01 10 11
int N, K;

int main() {
  scanf("%d%d", &N, &K);
  dp[1][1][0] = dp[1][2][1] = dp[1][2][2] = dp[1][1][3] = 1;
  for (reg int i = 2; i <= N; ++i) {
  	for (reg int j = 1; j <= i * 2; ++j) {
  	  dp[i][j][0] = (dp[i-1][j][0] % mod + dp[i-1][j][1] % mod + dp[i-1][j][2] % mod + dp[i-1][j-1][3]) % mod;
  	  dp[i][j][1] = (dp[i-1][j-1][0] % mod + dp[i-1][j][1] % mod + dp[i-1][j-2][2] % mod + dp[i-1][j-1][3] % mod) % mod;
  	  dp[i][j][2] = (dp[i-1][j-1][0] % mod+ dp[i-1][j-2][1] % mod + dp[i-1][j][2] % mod + dp[i-1][j-1][3] % mod) % mod;
  	  dp[i][j][3] = (dp[i-1][j-1][0] % mod + dp[i-1][j][1] % mod + dp[i-1][j][2] % mod + dp[i-1][j][3] % mod) % mod;
	}
  }
  printf("%lld
", (dp[N][K][0] % mod + dp[N][K][1] % mod + dp[N][K][2] % mod + dp[N][K][3] % mod) % mod);
  
  return 0;
}

(color{red}{mathcal{Source}})

(Educational Codeforces Round 51 [Rated for Div. 2])