[AHOI 2009]chess 中国象棋

Description

题库链接

给你一张 $N\times M$ 的棋盘。要求每行每列最多放两个棋子，问总方案数。

$1\leq N,M\leq 100$

Solution

记 $f_{i,j,k}$ 为前 $i$ 行还剩 $j$ 行可以放 $1$ 个棋子， $k$ 行放两个棋子的方案数。组合数学乱搞就好了。

Code

//It is made by Awson on 2018.3.17
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 100, yzh = 9999973;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(unsigned LL x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(unsigned LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int f[N+5][N+5][N+5], n, m, C[N+5][N+5];

void work() {
    for (int i = 0; i <= 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])%yzh;
    }
    read(n), read(m);
    f[0][0][m] = 1;
    for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m; j++)
        for (int k = 0; k <= m; k++)
        for (int p = 0; p <= 2; p++)
            for (int q = 0; q+p <= 2; q++)
                (f[i][j+q][k] += 1ll*f[i-1][j+p][k+q]*C[j+p][p]%yzh*C[k+q][q]%yzh) %= yzh;
    int ans = 0;
    for (int i = 0; i <= m; i++) for (int j = 0; j <= m; j++) (ans += f[n][i][j]) %= yzh;
    writeln(ans);
}
int main() {
    work(); return 0;
}