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;

}

Rikka

杭电多校第九场 HDU6415 Rikka with Nash Equilibrium dp

Rikka with Nash Equilibrium