[hdoj6415 Rikka with Nash Equilibrium][dp]

http://acm.hdu.edu.cn/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): 2021    Accepted Submission(s): 857

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

题意：将1 2 3....n*m填入一个n*m的矩阵中，要求最多有一个数既是它所在行的最大值又是其所在列的最大值，求方案数%k的值

题解：由于n*m肯定是其所在行和所在列的最大值，所以可知应该从n*m到1依次填数，保证当前所填数和之前填的数同行或者同列。dp[i][j][q]表示填完当前数之后已经有i行j列被填入数字，q=0表示当前的数填入的位置所在行之前没有被填充，q=1表示所在列之前没有被填充，q=2表示所在行和列都被填充了，可以得到转移方程(1)dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2])%k*(n*j-(i-1)*j)%k; (2)dp[i][j][1]=(dp[i][j-1][0]+dp[i][j-1][1]+dp[i][j-1][2])%k*(m*i-i*(j-1))%k; (3)dp[i][j][2]=(dp[i][j][0]+dp[i][j][1]+dp[i][j][2])%k*((i*j)-(q-1))%k;

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 ll dp[][][];
 int pre[][];
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         int n,m,k;
         scanf("%d%d%d",&n,&m,&k);
         memset(dp,,sizeof(dp));
         dp[][][]=n*m;
         for(int q=;q<=n*m;q++){
             for(int i=min(n,q);i>=;i--){
                 for(int j=min(m,q-i+);j>=;j--){
                     if(i*j<q-)break;
                     dp[i][j][]=(dp[i][j][]+dp[i][j][]+dp[i][j][])%k*((i*j)-(q-))%k;
                     dp[i][j][]=(dp[i-][j][]+dp[i-][j][]+dp[i-][j][])%k*(n*j-(i-)*j)%k;
                     dp[i][j][]=(dp[i][j-][]+dp[i][j-][]+dp[i][j-][])%k*(m*i-i*(j-))%k;
                 }
             }
         }
         printf("%lld\n",(dp[n][m][]+dp[n][m][]+dp[n][m][])%k);
     }
     return ;
 }

注意：这道题如果不通过判断某些条件及时跳出循环就会T掉