Description
Edward is the headmaster of Marjar University. He is enthusiastic about chess and often plays chess with his friends. What's more, he bought a large decorative chessboard with N rows and M columns.
Every day after work, Edward will place a chess piece on a random empty cell. A few days later, he found the chessboard was dominatedby the chess pieces. That means there is at least one chess piece in every row. Also, there is at least one chess piece in every column.
"That's interesting!" Edward said. He wants to know the expectation number of days to make an empty chessboard of N × M dominated. Please write a program to help him.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
There are only two integers N and M (1 <= N, M <= 50).
Output
For each test case, output the expectation number of days.
Any solution with a relative or absolute error of at most 10-8 will be accepted.
Sample Input
2 1 3 2 2
Sample Output
3.000000000000
2.666666666667
題意：
一個n行m列的棋盤，每次可以放一個棋子，問要使得棋盤的每行每列都至少有一個棋子 需要的放棋子次數的期望。
那麼對於每一顆棋子，在現有的棋盤上，它可能有四種影響：新佔了一行，新佔了一列，既佔了新的一行又佔了新的一列，無影響。
注意這裏的無影響指的不是下在同一個位置，這是不允許的，指的是已有(1,2)，(2,1)，下在(1,1)無影響，不增加行和列。
題解一：
dp[i][j][k] 已經佔據i行j列，走了k步的時候，還需要走的步數的期望。
#include<cstdio> #include<cstring> #include<iostream> using namespace std; double dp[55][55][55*55]; int main() { int t; scanf("%d",&t); while(t--) { int n,m; scanf("%d%d",&n,&m); memset(dp,0,sizeof(dp)); for(int i=n;i>=0;i--) for(int j=m;j>=0;j--) for(int k=i*j;k>=max(i,j);k--) { if(i==n&&j==m) continue; dp[i][j][k]+=1.0*(n-i)*j/(1.0*n*m-k)*dp[i+1][j][k+1]; dp[i][j][k]+=1.0*i*(m-j)/(1.0*n*m-k)*dp[i][j+1][k+1]; dp[i][j][k]+=1.0*(n-i)*(m-j)/(1.0*n*m-k)*dp[i+1][j+1][k+1]; dp[i][j][k]+=1.0*(i*j-k)/(1.0*n*m-k)*dp[i][j][k+1]; dp[i][j][k]+=1.0; } printf("%.12lf\n",dp[0][0][0]); } }
題解二：
dp[i][j][k]表示用了k個棋子共能佔領棋盤的i行j列的概率。
所以用dp[i][j][k]-dp[i][j][k-1]得到是第k個棋子恰好使得每行每列都佔領的概率。
#include<cstdio> #include<cstring> double dp[55][55][2550]; int n,m; int main() { int T,i,j,k; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); int sum=n*m; for(i=0;i<=n;i++) for(j=0;j<=m;j++) for(k=0;k<=sum;k++) dp[i][j][k]=0; dp[0][0][0]=1.0; for(k=1;k<=sum;k++) for(i=1;i<=n;i++) for(j=1;j<=m;j++) { dp[i][j][k]+=(dp[i][j][k-1]*(i*j-k+1)*1.0/(sum-k+1));//添加的位置沒有新增新行或新列 dp[i][j][k]+=(dp[i-1][j][k-1]*((n-i+1)*j)*1.0/(sum-k+1));//增加行不增加列 dp[i][j][k]+=(dp[i][j-1][k-1]*(m-j+1)*i*1.0/(sum-k+1));//增加列不增加行 dp[i][j][k]+=(dp[i-1][j-1][k-1]*(n-i+1)*(m-j+1)*1.0/(sum-k+1));//既增加列也增加行 // printf("i:%d j;%d k;%d dp:%lf\n",i,j,k,dp[i][j][k]); } double ans=0; for(k=1;k<=sum;k++) ans+=(dp[n][m][k]-dp[n][m][k-1])*k; printf("%.10lf\n",ans); } return 0; }