http://acm.hdu.edu.cn/showproblem.php?pid=6415
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; 1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 ll dp[85][85][3]; 5 int pre[85][85]; 6 int main() 7 { 8 int t; 9 scanf("%d",&t); 10 while(t--){ 11 int n,m,k; 12 scanf("%d%d%d",&n,&m,&k); 13 memset(dp,0,sizeof(dp)); 14 dp[1][1][0]=n*m; 15 for(int q=2;q<=n*m;q++){ 16 for(int i=min(n,q);i>=1;i--){ 17 for(int j=min(m,q-i+1);j>=1;j--){ 18 if(i*j<q-1)break; 19 dp[i][j][2]=(dp[i][j][0]+dp[i][j][1]+dp[i][j][2])%k*((i*j)-(q-1))%k; 20 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; 21 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; 22 } 23 } 24 } 25 printf("%lld\n",(dp[n][m][0]+dp[n][m][1]+dp[n][m][2])%k); 26 } 27 return 0; 28 } View Code
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
转载于:https://www.cnblogs.com/MekakuCityActor/p/10832756.html
