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;
}