CF1051D Bicolorings

CF1051D Bicolorings

Description

给定一个2×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

Solution

考虑状压每一列，然后就直接转移

#include<bits/stdc++.h>

using namespace std;

inline int read()
{
    int f = 1 ,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-') f = -1;
    }while(ch < '0'||ch > '9');
    do
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch >= '0'&&ch <= '9');
    return f*x;
}

const int MAXN = 1000 + 10;
const int MOD = 998244353;

int n;
int k;
long long dp[MAXN][MAXN * 2][4];

int main()
{
    n = read(),k = read();
    dp[1][1][0] = dp[1][2][1] = dp[1][2][2] = dp[1][1][3] = 1;
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<=min(k,i*2);j++)
        {
            dp[i][j][0] = (dp[i-1][j][0] + dp[i-1][j][1] + dp[i-1][j-1][3] + dp[i-1][j][2]) % MOD;
            dp[i][j][1] = (dp[i-1][j-1][0] + dp[i-1][j-2][2] + dp[i-1][j-1][3] + dp[i-1][j][1]) % MOD;
            dp[i][j][2] = (dp[i-1][j-1][0] + dp[i-1][j-2][1] + dp[i-1][j-1][3] + dp[i-1][j][2]) % MOD;
            dp[i][j][3] = (dp[i-1][j][3] + dp[i-1][j][1] + dp[i-1][j-1][0] + dp[i-1][j][2]) % MOD;
        }
    }
    cout << (dp[n][k][0] + dp[n][k][1] + dp[n][k][2] + dp[n][k][3]) % MOD << endl;
}