HDU

Rikka with Nash Equilibrium

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1460    Accepted Submission(s): 591

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

 

 

在一个矩阵中，如果仅有一个数同时是所在行所在列最大值，那么这个数满足纳什均衡。

构造一个n*m的矩阵，里面填入[1,n*m]互不相同的数字，求有多少种构造方案。

由题可得，满足纳什均衡的数一定是最大值n*m。因此我们可以从大往小依次将数填入矩阵，小数依附于大数的行或列，由此产生的三种行为：

1.所在列有大数，行+1 数+1

2.所在行有大数，列+1 数+1

3.所在行列都有大数，位于交界处，数+1

dp三种状态[占有行数][占有列数][占有个数]进行求解。因为本题数据很强，所以需要以下优化：

1.尽可能得减少取模次数

2.注意状态与遍历顺序保持一致

//记忆化搜索
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mod,n,m;
ll dp[81][81][6401];
ll dfs(ll x,ll y,ll z){
    if(dp[x][y][z]>-1) return dp[x][y][z];
    ll tmp=0;
    if(x<n) tmp+=y*(n-x)*dfs(x+1,y,z+1)%mod;
    if(y<m) tmp+=x*(m-y)*dfs(x,y+1,z+1)%mod;
    if(x*y>z) tmp+=(x*y-z)*dfs(x,y,z+1)%mod;
    return dp[x][y][z]=tmp;
}
int main()
{
    ll t;
    scanf("%lld",&t);
    while(t--){
        memset(dp,-1,sizeof(dp));  //答案有0的情况，因此初始化为-1
        scanf("%lld%lld%lld",&n,&m,&mod);
        dp[n][m][n*m]=1;
        ll ans=n*m*dfs(1,1,1)%mod;
        printf("%lld
",ans);
    }
}

//dp
#include<bits/stdc++.h>
#define MAX 82
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

ll dp[MAX][MAX][6402];    //注意顺序

int main()
{
    int t,n,m,MOD,i,j,k;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&MOD);
        memset(dp,0,sizeof(dp));
        dp[1][1][1]=n*m;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                for(k=1;k<n*m;k++){    //注意顺序
                    dp[i+1][j][k+1]+=(n-i)*j*dp[i][j][k];
                    if(dp[i+1][j][k+1]>=MOD) dp[i+1][j][k+1]%=MOD;
                    dp[i][j+1][k+1]+=(m-j)*i*dp[i][j][k];
                    if(dp[i][j+1][k+1]>=MOD) dp[i][j+1][k+1]%=MOD;
                    if(i*j>k){
                        dp[i][j][k+1]+=(i*j-k)*dp[i][j][k];
                        if(dp[i][j][k+1]>=MOD) dp[i][j][k+1]%=MOD;
                    }
                }
            }
        }
        printf("%lld
",dp[n][m][n*m]%MOD);
    }
    return 0;
}