? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?D. Painting The Wall

User ainta decided to paint a wall. The wall consists of?n2?tiles, that are arranged in an?n?×?n?table. Some tiles are painted, and the others are not. As he wants to paint it beautifully, he will follow the rules below.

Firstly user ainta looks at the wall. If there is at least one painted cell on each row and at least one painted cell on each column, he stops coloring. Otherwise, he goes to step 2.

User ainta choose any tile on the wall with uniform probability.

If the tile he has chosen is not painted, he paints the tile. Otherwise, he ignores it.

Then he takes a rest for one minute even if he doesn't paint the tile. And then ainta goes to step 1.

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However ainta is worried if it would take too much time to finish this work. So he wants to calculate the expected time needed to paint the wall by the method above. Help him find the expected time. You can assume that choosing and painting any tile consumes no time at all.

Input

The first line contains two integers?n?and?m?(1?≤?n?≤?2·103;?0?≤?m?≤?min(n2,?2·104)) — the size of the wall and the number of painted cells.

Next?m?lines goes, each contains two integers?ri?and?ci?(1?≤?ri,?ci?≤?n) — the position of the painted cell. It is guaranteed that the positions are all distinct. Consider the rows of the table are numbered from?1?to?n. Consider the columns of the table are numbered from1?to?n.

Output

In a single line print the expected time to paint the wall in minutes. Your answer will be considered correct if it has at most?10?-?4?absolute or relative error.

Sample test(s) input 5 2
2 3
4 1 output 11.7669491886 input 2 2
1 1
1 2 output 2.0000000000 input 1 1
1 1 output 0.0000000000

題意：有一個n*n的墻，現(xiàn)在小明來刷墻，如果每一行每一列都至少有一個格子刷過了就停止工作，否則每次隨機選一個格子，如果刷過了就不刷如果沒刷過就刷，然后休息一分鐘，求停止工作時時間的數(shù)學(xué)期望（開始之前已經(jīng)有m個格子刷過了）
題解：dp[i][j]表示還有i行j列未刷
初始化： dp[i][0]=((n-i)/n)*dp[i][0]+dp[i-1][0]*i/n+1;
dp[0][j]=((n-j)/n)*dp[0][j]+dp[0][j-1]*j/n+1;
轉(zhuǎn)移： dp[i][j]=dp[i][j]*(n-i)(n-j)/n^2+dp[i-1][j]*(i*(n-j))/n^2+dp[i][j-1]*((n-i)*j)/n^2+dp[i-1][j-1]*(i*j)/n^2+1;

#include<iostream> #include<cstdio> using namespace std; double dp[2010][2010]; int n,m,a[2010],b[2010]; int main() {cin>>n>>m;int x,y;int l=n,r=n;for(int i=0; i<m; i++){cin>>x>>y;if(!a[x]) l--;if(!b[y]) r--;a[x]=1,b[y]=1;}for(int i=1; i<=n; i++) dp[i][0]=dp[i-1][0]+(double)n/i;for(int j=1; j<=n; j++) dp[0][j]=dp[0][j-1]+(double)n/j;for(int i=1; i<=n; i++){for(int j=1; j<=n; j++){dp[i][j]=(dp[i-1][j]*i*(n-j)+n*n+dp[i][j-1]*j*(n-i)+dp[i-1][j-1]*i*j)/(n*n-(n-i)*(n-j));}}printf("%0.10f\n",dp[l][r]);return 0; } 代碼

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轉(zhuǎn)載于:https://www.cnblogs.com/zxhl/p/4846283.html

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