HDU6415：Rikka with Nash Equilibrium（dp）

Problem Description

Nash Equilibrium is an important concept in game theory.

Rikka and Yuta are playing a simple matrix game. At the beginning of the game, Rikka shows an n×m integer matrix A. And then Yuta needs to choose an integer in [1,n], Rikka needs to choose an integer in [1,m]. Let i be Yuta's number and j be Rikka's number, the final score of the game is Ai,j. 

In the remaining part of this statement, we use (i,j) to denote the strategy of Yuta and Rikka.

For example, when n=m=3 and matrix A is 

⎡⎣⎢111241131⎤⎦⎥

If the strategy is (1,2), the score will be 2; if the strategy is (2,2), the score will be 4.

A pure strategy Nash equilibrium of this game is a strategy (x,y) which satisfies neither Rikka nor Yuta can make the score higher by changing his(her) strategy unilaterally. Formally, (x,y) is a Nash equilibrium if and only if:

{Ax,y≥Ai,y  ∀i∈[1,n]Ax,y≥Ax,j  ∀j∈[1,m]

In the previous example, there are two pure strategy Nash equilibriums: (3,1) and (2,2).

To make the game more interesting, Rikka wants to construct a matrix A for this game which satisfies the following conditions:
1. Each integer in [1,nm] occurs exactly once in A.
2. The game has at most one pure strategy Nash equilibriums. 

Now, Rikka wants you to count the number of matrixes with size n×m which satisfy the conditions.

 

 

Input

The first line contains a single integer t(1≤t≤20), the number of the testcases.

The first line of each testcase contains three numbers n,m and K(1≤n,m≤80,1≤K≤109).

The input guarantees that there are at most 3 testcases with max(n,m)>50.

 

 

Output

For each testcase, output a single line with a single number: the answer modulo K.

 

 

Sample Input

2 3 3 100 5 5 2333

 

 

Sample Output

64 1170

題意：給個n*m矩陣，用1~n*m填進去，使得Nash equilibriums至多一個，Nash equilibriums值某個數在所在行和列是最大的。

思路：從大到小填進去，dp[i][j][k]表示填了i個數佔了j行k列的方案數，顯然新填一個數要麼j+1要麼k+1要麼都不加，不可能j和k都加，然後遞推下去，加一些判斷優化更快。

# include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 6403;
LL dp[maxn][81][81], mod;
inline void add(LL &a, LL b){
    a += b;
    if(a > mod) a -= mod;
}
int main()
{
    int T, n, m;
    for(scanf("%d",&T);T;--T){
        scanf("%d%d%lld",&n,&m,&mod);
        memset(dp, 0, sizeof(dp));
        dp[1][1][1] = n*m;
        for(int i=1; i<n*m; ++i){
            for(int j=1; j<=n; ++j){
                for(int k=1; k<=m; ++k){
                    if(max(j,k)>i) break;
                    if(!dp[i][j][k]) continue;
                    if(j<n) add(dp[i+1][j+1][k], dp[i][j][k]*k%mod*(n-j)%mod);
                    if(k<m) add(dp[i+1][j][k+1], dp[i][j][k]*j%mod*(m-k)%mod);
                    if(i<j*k) add(dp[i+1][j][k],dp[i][j][k]*(j*k-i)%mod);
                }
            }
        }
        printf("%lld\n",dp[n*m][n][m]);
    }
}