dp

dp[k][i][j]表示放k个棋子用了i行j列的方法数。

从小到大放，每次必须放在已经放过的行或列上，否则会出现多个纳什均衡点。

转移方式有三种，新放一行，新放一列，放在已经放过的行和列。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int X = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();    return w ? -X : X;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 105;int _, n, m, p;ll dp[81*81][81][81];int main(){    for(_ = read(); _; _ --){        n = read(), m = read(), p = read();        full(dp, 0);        dp[1][1][1] = n * m;        for(int k = 2; k <= n * m; k ++){            for(int i = 1; i <= n; i ++){                for(int j = 1; j <= m; j ++){                    if(k > i * j) continue;                    dp[k][i][j] += 1LL * dp[k - 1][i - 1][j] * j % p * (n - i + 1) % p;                    dp[k][i][j] += 1LL * dp[k - 1][i][j - 1] * i % p * (m - j + 1) % p;                    dp[k][i][j] += 1LL * dp[k - 1][i][j] % p * (i * j - k + 1) % p;                }            }        }        printf("%lld\n", dp[n*m][n][m]);    }    return 0;}