CF1051D Bicolorings dp

水题一道

 

$f[i][j][S]$表示$2 * i$的矩形，有$j$个联通块，某尾状态为$S$

然后转移就行了...

 

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define re register
    #define de double
    #define le long double
    #define ri register int
    #define ll long long
    #define sh short
    #define pii pair<int, int>
    #define mp make_pair
    #define pb push_back
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)    
    extern inline char gc() {
        static char RR[23456], *S = RR + 23333, *T = RR + 23333;
        if(S == T) fread(RR, 1, 23333, stdin), S = RR;
        return *S ++;
    }
    inline int read() {
        int p = 0, w = 1; char c = gc();
        while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }
        while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();
        return p * w;
    }
    int wr[50], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('0');
        if(o < 0) o = -o, pc('-');
        while(o) wr[++ rw] = o % 10, o /= 10;
        while(rw) pc(wr[rw --] + '0');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; } 
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b, 1 : 0; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b, 1 : 0; }
}
using namespace std;
using namespace remoon;

#define mod 998244353

inline void inc(ll &a, ll b) {
    a += b; if(a >= mod) a %= mod;
}

int n, k;
ll f[1005][2010][4];

int main() {
    n = read(); k = read();
    f[1][1][0] = 1; f[1][2][2] = 1;
    f[1][2][1] = 1; f[1][1][3] = 1;
    rep(i, 2, n)
    rep(j, 1, i << 1) {
        inc(f[i][j][0], f[i - 1][j][0] + f[i - 1][j][1] + f[i - 1][j][2] + f[i - 1][j - 1][3]);
        inc(f[i][j][1], f[i - 1][j - 1][0] + f[i - 1][j][1] + f[i - 1][j - 2][2] + f[i - 1][j - 1][3]);
        inc(f[i][j][2], f[i - 1][j - 1][0] + f[i - 1][j - 2][1] + f[i - 1][j][2] + f[i - 1][j - 1][3]);
        inc(f[i][j][3], f[i - 1][j - 1][0] + f[i - 1][j][1] + f[i - 1][j][2] + f[i - 1][j][3]);
    }
    ll ans = 0;
    inc(ans, f[n][k][0] + f[n][k][1] + f[n][k][2] + f[n][k][3]);
    write(ans);
    return 0;
}