[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;

}