codeforces 1051 D. Bicolorings (DP)

D. Bicolorings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $$$998244353$$$.

Input

The only line contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 1000$$$, $$$1 \le k \le 2n$$$) — the number of columns in a grid and the number of components required.

Output

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

Examples

Input

3 4

Output

12

Input

4 1

Output

2

Input

1 2

Output

2

Note

One of possible bicolorings in sample $$$1$$$:

题意

给一个2行n列的矩阵填上黑色和白色，求连通块个数为k个的填色方案数量(mod 998244353)

分析

因为只有两行，为n-1列的矩阵增加1列的情况数只有很少，容易想到用 $$$(i, k)$$$ 表示 $$$i$$$ 列有 $$$k$$$ 个连通块的矩阵， 但是它在向 $$$i+1$$$ 列的矩阵转移时，需要知道最后一列的状态，所以可以用 $$$0$$$, $$$1$$$, $$$2$$$, $$$3$$$表示最后一列为 $$$00$$$, $$$01$$$, $$$10$$$, $$$11$$$，那么状态就增加一维变成 $$$(i, k, s)$$$，然后就是分析递推关系：

$$$(i,k,0)$$$ 的矩阵，可以由 $$$i-1$$$ 列的矩阵添加一列 $$$00$$$ 得到，当它的结尾为 $$$00$$$, $$$01$$$, $$$10$$$, $$$11$$$时，分别会让连通块个数：不变，不变，不变，+1,所以 $$$(i,k,0)$$$由 $$$(i-1,k,0)$$$, $$$(i-1,k,1)$$$, $$$(i-1,k,2)$$$, $$$(i-1,k-1,3)$$$得到： $$$$$$ \begin{align} dp[i][k][0,0]=~~~& dp[i-1][k][0,0]\\ +& dp[i-1][k][0,1]\\ +& dp[i-1][k][1,0]\\ +& dp[i-1][k-1][1,1] \end{align} $$$$$$ $$$(i,k,1)$$$的矩阵同理，为$$$i-1$$$列的矩阵添加 $$$01$$$，当结尾为 $$$00$$$, $$$01$$$, $$$10$$$, $$$11$$$时，分别会使连通块的个数：+1，不变，+2，+1，所以$$$(i,k,1)$$$由$$$(i-1,k-1,0)$$$,$$$(i-1,k,1)$$$,$$$(i-1,k-2,2)$$$,$$$(i-1,k-1,3)$$$得到： $$$$$$ \begin{align} dp[i][k][0,1]=~~~& dp[i-1][k-1][0,0]\\ +& dp[i-1][k][0,1]\\ +& dp[i-1][k-2][1,0]\\ +& dp[i-1][k-1][1,1] \end{align} $$$$$$ (i,k,2)同理可得： $$$$$$ \begin{align} dp[i][k][1,0]=~~~& dp[i-1][k-1][0,0]\\ +& dp[i-1][k-2][0,1]\\ +& dp[i-1][k][1,0]\\ +& dp[i-1][k-1][1,1] \end{align} $$$$$$ (i,k,3)同理可得： $$$$$$ \begin{align} dp[i][k][1,1]=~~~& dp[i-1][k-1][0,0]\\ +& dp[i-1][k][0,1]\\ +& dp[i-1][k][1,0]\\ +& dp[i-1][k][1,1] \end{align} $$$$$$ 于是得到了完整的递推公式，只需要从下面的状态开始， $$$$$$ \begin{align} dp[1][1][0,0]=1\\ dp[1][2][0,1]=1\\ dp[1][2][1,0]=1\\ dp[1][1][1,1]=1 \end{align} $$$$$$ 就能推到出所有的状态，最后对dp[n][k]的所有情况求和就是答案了。

注意当k为1时，是不存在k-2的状态的，需要特判一下避免超出数组范围

总结

动态规划的状态定义很关键，必须抓住状态之间的联系；递推式的推导也需要深入思考

代码

 

#include<stdio.h>
typedef long long LL;
#define mod 998244353
int dp[1003][2006][4] = {0};
int main() {
    int n, lm;
    scanf("%d %d", &n, &lm);
    //初始化
    dp[1][1][0] = 1;//00
    dp[1][2][2] = 1;//10
    dp[1][2][1] = 1;//01
    dp[1][1][3] = 1;//11
    LL temp=0;
    for (int i = 2; i <= n; ++i) {
        for (int k = 1; k <= (i << 1); ++k) {
            temp = 0;//使用temp求和来避免溢出
            temp =temp
                + dp[i - 1][k][1]//01
                + dp[i - 1][k][0]//00
                + dp[i - 1][k][2]//10
                + dp[i - 1][k - 1][3];//11
            dp[i][k][0] = temp % mod;
            temp = 0;
            temp = temp 
                + dp[i - 1][k][1]//01
                + dp[i - 1][k-1][0]//00
                + (k>=2?dp[i - 1][k - 2][2]:0)//10
                + dp[i - 1][k-1][3];//11
            dp[i][k][1] = temp%mod;
            temp = 0;
            temp = temp 
                + (k>=2?dp[i - 1][k - 2][1]:0)//01
                + dp[i - 1][k-1][0]//00
                + dp[i - 1][k][2]//10
                + dp[i - 1][k-1][3];//11
            dp[i][k][2] = temp%mod;
            temp = 0;
            temp = temp 
                + dp[i - 1][k][1]//01
                + dp[i - 1][k - 1][0]//00
                + dp[i - 1][k][2]//10
                + dp[i - 1][k][3];//11
            dp[i][k][3] = temp%mod;
            temp = 0;
        }
    }
    LL ans = 0;
    ans = ans + dp[n][lm][0] + dp[n][lm][1] + dp[n][lm][2] + dp[n][lm][3];
    ans = ans%mod;
    printf("%I64d\n", ans);
}