题目链接

Solution [AHOI2009]中国象棋

题目大意:求在\(n\)行\(m\)列的棋盘上放置若干个炮使得它们互不攻击的方案数,对\(9999973\)取模

分析:没有炮互相攻击等价于没有三个炮在同一行/列

考虑用\(f[i][m_1][m_2]\)表示前\(i\)行,有\(m_1\)列有一个炮,\(m_2\)列有两个炮的方案数

可以用刷表法降低思维难度

初始条件:

\(f[1][0][0] = 1\)

\(f[1][1][0] = m\)

\(f[1][2][0] = C_m^2\)

这个比较简单,转移可以大力分类讨论,即讨论第\(i + 1\)行的放置个数,以及放置方案

• \(1.i + 1\)行放\(0\)个

\(f[i + 1][m_1][m_2] += f[i][m_1][m_2]\)

• \(2.i + 1\)行放\(1\)个,并且放在空列(有\(0\)个棋子的列)

\(f[i + 1][m_1 + 1][m_2] += f[i][m_1][m_2] \times (m - m_1 - m_2)\)

• \(3.i + 1\)行放\(1\)个,并且放在有\(1\)个棋子的列

\(f[i + 1][m_1 - 1][m_2 + 1] += f[i][m_1][m_2] \times m_1\)

• \(4.i + 1\)行放\(2\)个,都放在空列

\(f[i + 1][m_1 + 2][m_2] += f[i][m_1][m_2] \times C_{m-m_1-m_2}^2\)

• \(5.i + 1\)行放\(2\)个,一个放在空列,一个放在有\(1\)个棋子的列

\(f[i + 1][m_1][m_2 + 1] += f[i][m_1][m_2] \times m_1 \times (m - m_1 - m_2)\)

• \(6.i + 1\)行放\(2\)个,都放在有\(1\)个棋子的列

\(f[i + 1][m_1 - 2][m_2 + 2] += f[i][m_1][m_2] \times C_{m_1}^2\)

然后注意一下边界条件,避免非法状态和非法转移即可

#include <cstdio>
using namespace std;
typedef long long ll;
const int maxn = 128,mod = 9999973;
int n,m;
ll f[maxn][maxn][maxn],ans;
inline ll C2(ll x){return x * (x - 1) / 2;}
int main(){
    scanf("%d %d",&n,&m);
    f[1][0][0] = 1;
    f[1][1][0] = m;
    f[1][2][0] = C2(m);
    for(int i = 1;i < n;i++)
        for(int m1 = 0;m1 <= m;m1++)
            for(int m2 = 0;m1 + m2 <= m;m2++){
                f[i + 1][m1][m2] += f[i][m1][m2],f[i + 1][m1][m2] %= mod;
                if(m1 + 1 <= m && m1 + m2 + 1 <= m)f[i + 1][m1 + 1][m2] += f[i][m1][m2] * ll(m - m1 - m2),f[i + 1][m1 + 1][m2] %= mod;
                if(m1 >= 1 && m2 < m)f[i + 1][m1 - 1][m2 + 1] += f[i][m1][m2] * (ll)m1,f[i + 1][m1 - 1][m2 + 1] %= mod;
                if(m1 + 2 <= m && m1 + m2 + 2 <= m)f[i + 1][m1 + 2][m2] += f[i][m1][m2] * C2(m - m1 - m2),f[i + 1][m1 + 2][m2] %= mod;
                if(m2 + 1 <= m && m1 + m2 + 1 <= m)f[i + 1][m1][m2 + 1] += f[i][m1][m2] * ll(m - m1 - m2) * m1,f[i + 1][m1][m2 + 1] %= mod;
                if(m1 >= 2 && m2 + 2 <= m)f[i + 1][m1 -  2][m2 + 2] += f[i][m1][m2] * C2(m1),f[i + 1][m1 - 2][m2 + 2] %= mod;
            }
    for(int m1 = 0;m1 <= m;m1++)
        for(int m2 = 0;m1 + m2 <= m;m2++)
            ans += f[n][m1][m2],ans %= mod;
    printf("%lld\n",ans);
    return 0;
}

题解 P2051 【[AHOI2009]中国象棋】

原文：https://www.cnblogs.com/colazcy/p/11520079.html