HDU - 6122 Color the chessboard
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我们能分析出来不可能列和行有连续一样的数字, 所以我们能得出图形是由ABABAB 和 BABABA 这两种形式的东西叠在一起构成的,
要么是一列一列叠出来, 要么一行一行叠出来, 最后减去重复的, 也就是所有相邻数字都不一样的。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 1e3 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) a += b; if(a >= mod) a -= mod; template<class T, class S> inline void sub(T &a, S b) a -= b; if(a < 0) a += mod; template<class T, class S> inline bool chkmax(T &a, S b) return a < b ? a = b, true : false; template<class T, class S> inline bool chkmin(T &a, S b) return a > b ? a = b, true : false; mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n, m, r[N], c[N], a[N]; char s[N][N]; int calc(int n) int p = 0; for(int i = 1; i <= n; i++) if(a[i] == -1) continue; p = i; break; if(!p) return 2; int now = (p & 1) ? a[p] : (a[p] ^ 1); for(int i = 1; i <= n; i++) if(~a[i] && now != a[i]) return 0; now ^= 1; return 1; int main() int T; scanf("%d", &T); while(T--) scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%s", s[i] + 1); for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(s[i][j] == ‘R‘) a[j] = 0; else if(s[i][j] == ‘B‘) a[j] = 1; else a[j] = -1; r[i] = calc(m); for(int j = 1; j <= m; j++) for(int i = 1; i <= n; i++) if(s[i][j] == ‘R‘) a[i] = 0; else if(s[i][j] == ‘B‘) a[i] = 1; else a[i] = -1; c[j] = calc(n); int wayr = 1; int wayc = 1; int wayrc = 0; int f0 = true, f1 = true; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(s[i][j] == ‘?‘) continue; int a = (i - 1) & 1; int b = (j - 1) & 1; int c = s[i][j] == ‘R‘ ? 0 : 1; if((a ^ b ^ c) != 0) f0 = false; if((a ^ b ^ c) != 1) f1 = false; wayrc = f0 + f1; for(int i = 1; i <= n; i++) wayr = 1LL * wayr * r[i] % mod; for(int i = 1; i <= m; i++) wayc = 1LL * wayc * c[i] % mod; printf("%d\n", ((wayr + wayc) % mod - wayrc + mod) % mod); return 0; /* */
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