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
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:
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.
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.
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
Recommend
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;
}