Zoj -- 3822 -- probability DP of the second question

This question is more difficult than the previous probability DP...

And the source of this question is the last time I went to the Mudanjiang semi-finals ..

FML ------ don't want to talk more

DP [I] [J] [k] indicates that K pieces occupy the J column of the I row. <each coordinate is (x, y) the pawns can all occupy column Y in row x>

So if K pieces have been placed and the J column of the I row is occupied, then for the (k + 1) there will be four different cases for each piece. <ensure that there are spaces on the board while they can be placed again>

Here, N * m-K refers to the number of N * m grids that can be placed with K grids.

1.-The k + 1 piece is placed in a column in the J column of the occupied I row.

You can find p1 = (I * j-k)/(n * m-K)

The corresponding state transition equation is DP [I] [J] [k] = DP [I] [J] [k-1] * P1;

2.-This k + 1 piece is placed in one of the occupied J columns but not in one of the occupied I rows.

We can find P2 = (n-I) * j/(n * m-K)

Corresponding to DP [I + 1] [J] [k] = DP [I] [J] [k-1] * P2; // This piece occupies more than one line

3-This k + 1 piece is placed in one of the occupied I rows but not in one of the occupied J columns.

You can obtain P3 = (m-j) * I/(n * m-K)

Corresponding to DP [I] [J + 1] [k] = DP [I] [J] [k-1] * P3; // This piece occupies one more column

4-This k + 1 piece is placed on the unoccupied (n-I) Row and (m-j) column.

Obtain P4 = (n-I) * (m-j)/(n * m-K)

Corresponding to DP [I + 1] [J + 1] [k] = DP [I] [J] [k-1] * P4; // This piece occupies one more row and one column.

In fact, we can find that when n> = 1 & M> = 1, the minimum number of pieces to be placed is 1. the maximum number of pieces to be placed is maxnum = (max (n, m)-1) * min (n, m) + 1

It is better to add sum.

Ans = sigma (DP [N] [m] [k]) k = 1, 2 ,......... Maxnum;

 1 #include <iostream> 2 #include <cstring> 3 #include <iomanip> 4 #include <algorithm> 5 using namespace std; 6  7 const int size = 55; 8 double dp[size][size][size*size]; 9 10 int main()11 {12     int t , n ,m , maxNum;13     double ans;14     cin >> t;15     while( t-- )16     {17         ans = 0;18         memset( dp , 0 , sizeof(dp) );19         dp[0][0][0] = 1;20         cin >> n >> m;21         maxNum = ( max(n,m)-1 ) * min(n,m) + 1;22         for( int i = 1 ; i<=n ; i++ )23         {24             for( int j = 1 ; j<=m ; j++ )25             {26                 for( int k = 1 ; k<=maxNum ; k++ )27                 {28                     dp[i][j][k] = ( dp[i-1][j][k-1] * (n-i+1) * j  + dp[i][j-1][k-1] * (m-j+1) * i  + dp[i-1][j-1][k-1] * (n-i+1) * (m-j+1) ) / ( n*m-k+1 );29                     if( i==n && j==m )30                         continue;31                     else32                         dp[i][j][k] += dp[i][j][k-1] * ( i*j-k+1 ) / (n*m-k+1);33                 }34             }35         }36         for( int k = 1 ; k<=maxNum ; k++ )37         {38             ans += k * dp[n][m][k];39         }40         cout << setiosflags(ios::fixed);41         cout << setprecision(12) << ans <<endl;42     }43     return 0;44 }

View code

 

Today:

Why do we always expect something to come?

Because we artificially increase the probability of occurrence or even approximate 100%.

But what we fear always comes.

Because of our own fear, we are unable to make full use of our ordinary standards.

 

Zoj -- 3822 -- probability DP of the second question