【BZOJ3294/洛谷3158】[CQOI2011]放棋子（组合数+DP）

题目：

洛谷3158

分析：

某OIer兔崽子的此题代码中的三个函数名：dfs、ddfs、dddfs（充满毒瘤的气息

显然，行与行之间、列与列之间是互相独立的。考虑背包，用\(f[k][i][j]\)表示用前\(k\)种颜色占了\(i\)行\(j\)列的方案数，\(g[i][j]\)表示用颜色\(k\)占据\(i\)行\(j\)列的方案数，\(c[i]\)表示颜色为\(i\)的棋子数，就有如下方程：

\[f[k][i][j]=\sum _{a=0}^i \sum_{b=0}^j f[k-1][i-a][j-b]\times g[a][b]\times C_{n-(i-a)}^a\times C_{m-(j-b)}^b(ab\geq c[i]) \]

\(g[i][j]\)在算的时候注意要减去有空行或空列的情况（枚举有多少行、列不是空的）。注意要\(a=i\)且\(b=j\)的情况要跳过：

\[g[i][j]=C_{ij}^{c[k]}-\sum_{a=0}^i \sum_{b=0}^j g[a][b]\times C_i^a \times C_j^b(ij\geq c[k]) \]

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
using namespace std;

namespace zyt
{
	template<typename T>
	inline void read(T &x)
	{
		char c;
		bool f = false;
		x = 0;
		do
			c = getchar();
		while (c != '-' && !isdigit(c));
		if (c == '-')
			f = true, c = getchar();
		do
			x = x * 10 + c - '0', c = getchar();
		while (isdigit(c));
		if (f)
			x = -x;
	}
	template<typename T>
	inline void write(T x)
	{
		static char buf[20];
		char *pos = buf;
		if (x < 0)
			putchar('-'), x = -x;
		do
			*pos++ = x % 10 + '0';
		while (x /= 10);
		while (pos > buf)	
			putchar(*--pos);
	}
	typedef long long ll;
	const int N = 40, T = 20, p = 1e9 + 9;
	int n, m, c, arr[T], C[N * N][N * N], f[T][N][N], g[N][N];
	void init()
	{
		for (int i = 0; i < N * N; i++)
		{
			C[i][0] = 1;
			for (int j = 1; j <= i; j++)
				C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % p;
		}
	}
	int work()
	{
		init();
		read(n), read(m), read(c);
		for (int i = 1; i <= c; i++)
			read(arr[i]);
		f[0][0][0] = 1;
		for (int k = 1; k <= c; k++)
		{
			memset(g, 0, sizeof(g));
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= m; j++)
				{
					if (i * j >= arr[k])
					{
						g[i][j] = C[i * j][arr[k]];
						for (int a = 0; a <= i; a++)
							for (int b = 0; b <= j; b++)
							{
								if (a != i || b != j)
									g[i][j] = (g[i][j] - (ll)g[a][b] * 
										C[i][a] % p * C[j][b] % p + p) % p;
							}
					}
				}
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= m; j++)
					for (int a = 1; a <= i; a++)
						for (int b = 1; b <= j; b++)
							if (a * b >= arr[k])
								f[k][i][j] = (f[k][i][j] + 
									(ll)f[k - 1][i - a][j - b] * g[a][b] % p * 
									C[n - (i - a)][a] % p * C[m - (j - b)][b]) % p;
		}
		int ans = 0;
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= m; j++)
				ans = (ans + f[c][i][j]) % p;
		write(ans);
		return 0;
	}
}
int main()
{
	return zyt::work();
}