showproblem.php?pid=6415

Rikka with Nash Equilibrium

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1524 Accepted Submission(s): 626

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

3 3 100

5 5 2333

Sample Output

1170

Source

2018 Multi-University Training Contest 9

题意：

纳什平衡点，所在行最大，列也是最大的， n*m的矩阵内只有一个纳什平衡点，

问 1-n*m的数，只出现一次，并且数字只在 1-n*m 内，问有多少种放法满足条件，

思路：

采用 dp 方法， dp[x][y][num] 代表在第 x 行， y列放置数字为num 有多少中方法，

最大的数一定是 n*m ，有 n*m中方法，倒数第二大 n*m-1 只能和他在一行或在一列，否则是不行的，有 n+m-1种方法。

倒数第三大的数字 n-m-2 根据上一个数字，如果 n*m-1 和n*m 在一行，那么这个数就有 n+2*m-2 如果在一列就有 2*n+m-2;

依次往下，一直到最后一个数。

如果划分区域的话，把矩阵划分成 4 块，上下左右，

代码：

#include <bits/stdc++.h>
#include <stdio.h>
#define rep(i,a,n) for(int i = a;i<=n;i++)

typedef long long ll;

const int maxn = 82;
using namespace std;

int n,m,mod;
ll dp[maxn][maxn][maxn*maxn];

int main(int argc, char const *argv[])
{
    int T;
    scanf("%d",&T);
    while(T--)
    {    
        scanf("%d %d %d",&n,&m,&mod);
        dp[n][m][n*m] = 1;
        for(ll num = n*m -1 ; num >=1 ;num --)
        {
            for(int i = n; i >=1 ;i--)
            {
                for(int j = m; j>=1 ;j--)
                {
                    if( i*j < num) break; // left up corner  
                    dp[i][j][num] = j*(n-i)%mod*dp[i+1][j][num+1]%mod; // left down corner
                    dp[i][j][num] = (dp[i][j][num]  +  i*(m-j)%mod*dp[i][j+1][num+1] )%mod; // right up corner
                    dp[i][j][num] = (dp[i][j][num]  +  (i*j-num)%mod*dp[i][j][num+1])%mod;  // left down corner
                }
            }
        }
        printf("%lld\n",(n*m%mod*dp[1][1][1])%mod);
    }
    return 0;
}

@HDU - 6415 - 杭电多校第九场 - Rikka with Nash Equilibrium (DP )

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=6415

Rikka with Nash Equilibrium

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