CF1237F Balanced Domino Placements

Solution

显然可以先考虑横着的骨牌，再考虑竖着的骨牌。但是思路卡在了选取横着的骨牌会对竖着的骨牌的相邻对数产生影响。

然而事实上我们只需要换一个统计顺序，先考虑横着的骨牌的列和竖着的骨牌的行，然后再考虑横着的骨牌的行和竖着的骨牌的列（因为单行/列没有双行/列的必须相邻的限制，因此可以简单地通过剩余行/列数统计）。

前面一项可以简单地$dp$得到，后面一项组合计数即可。

时间复杂度$O(n^2)$。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondtypedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI; const lod eps = 1e-11;
const lod pi = acos(-1);
const int mods = 998244353;
const int oo = 1 << 30;
const ll loo = 1ll << 62;
const int MAXN = 4005;
const int INF = 0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read() {int f = 1, x = 0; char c = getchar();while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }while (c >= '0' && c <= '9') { x = (x << 3) + (x << 1) + (c ^ 48); c = getchar(); }return x * f;
}
int tagx[MAXN], tagy[MAXN], fn[MAXN][MAXN], fm[MAXN][MAXN], fac[MAXN], inv[MAXN];
int upd(int x , int y) { return x + y >= mods ? x + y - mods : x + y; }
int C(int x, int y) { return x < y ? 0 :1ll * fac[x] * inv[y] % mods * inv[x - y] % mods; }
int quick_pow(int x, int y) {int ret = 1;for(; y ; y >>= 1) {if (y & 1) ret = 1ll * ret * x % mods;x = 1ll * x * x % mods;}return ret;
}
void Init(int n) {fac[0] = 1;for (int i = 1; i <= n ; ++ i) fac[i] = 1ll * fac[i - 1] * i % mods;inv[n] = quick_pow(fac[n], mods - 2);for (int i = n - 1; i >= 0 ; -- i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mods;
}
signed main() {int n = read(), m = read(), k = read();Init(max(n, m));for (int i = 1; i <= k ; ++ i) tagx[read()] = 1, tagy[read()] = 1, tagx[read()] = 1, tagy[read()] = 1;fn[0][0] = 1;for (int i = 1; i <= n ; ++ i)for (int j = 0; j * 2 <= i; ++ j) fn[i][j] = (i > 1 && !tagx[i] && !tagx[i - 1]) ? upd(fn[i - 1][j], fn[i - 2][j - 1]) : fn[i - 1][j];fm[0][0] = 1;for (int i = 1; i <= m ; ++ i)for (int j = 0; j * 2 <= i; ++ j) fm[i][j] = (i > 1 && !tagy[i] && !tagy[i - 1]) ? upd(fm[i - 1][j], fm[i - 2][j - 1]) : fm[i - 1][j];int rn = 0, rm = 0;for (int i = 1; i <= n ; ++ i) rn += !tagx[i];for (int i = 1; i <= m ; ++ i) rm += !tagy[i];int ans = 0;for (int i = 0; i * 2 <= rn ; ++ i)for (int j = 0; j * 2 <= rm; ++ j) ans = upd(ans, 1ll * fac[i] * fac[j] % mods * C(rm - j * 2, i) % mods * C(rn - i * 2, j) % mods * fn[n][i] % mods * fm[m][j] % mods);printf("%d\n", ans);return 0;
}