Rikka with Nash Equilibrium

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1251    Accepted Submission(s): 506

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

Source

2018 Multi-University Training Contest 9

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题意：

在一个矩阵中，如果某一个数字是该行该列的最大值，则这个数满足纳什均衡。

要求构造一个n*m的矩阵，里面填的数字各不相同且范围是【1，m*n】，且矩阵内最多有一个数满足纳什平衡，问有多少种构造方案。

分析：

从大到小往矩阵里填数，则填的数会多占领一行或者多占领一列或者不占领（上方左方都有比他更大的数）

多占领一行，则这一行可任意填的位置是是这一行还没填的列

多占领一列，同理

特殊考虑：有更大的数还没填进去的情况

参考博客：

https://blog.csdn.net/monochrome00/article/details/81875980

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
//const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll n, m, mod, dp[85][85][85*85];
int main() {
    ios::sync_with_stdio(0);
    ll t;
    cin >> t;
    while( t -- ) {
        cin >> n >> m >> mod;
        dp[n][m][n*m] = 1;  //占领了n-n+1行m-m+1列，放入了n*m-n*m+1个数字
        for( ll k = n*m-1; k >= 1; k -- ) {
            for( ll i = n; i >= 1; i -- ) { //从最后一行一列开始放最大的数字
                for( ll j = m; j >= 1; j -- ) {
                    if( i*j < k ) {
                        break;
                    }
                    dp[i][j][k] = j*(n-i)%mod*dp[i+1][j][k+1]%mod; //多占领了一行，这一行还没放的位置可以随意放
                    dp[i][j][k] = (dp[i][j][k]+i*(m-j)%mod*dp[i][j+1][k+1]%mod)%mod; //多占领了一列，同上
                    dp[i][j][k] = (dp[i][j][k]+(i*j-k)%mod*dp[i][j][k+1]%mod)%mod; //还有更大的数没有放进去的情况
                }
            }
        }
        cout << n*m%mod*dp[1][1][1]%mod << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/l609929321/p/9512338.html