Painting The Wall CodeForces

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.

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 ≤ 2·10^3; 0 ≤ m ≤ min(n2, 2·10^4)) — 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 from 1 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.

Example
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

大致题意：一个人去刷n*n规格的墙，每次花费一个单位时间随机的选一个格子染色，如果每行每列至少都有一个格子被染色时停止，一开始已经有一些格子被染色，问你停止染色时的时间期望是多少。

思路：假设dp[i][j]表示此时已经有i行j列满足条件，还需花费的时间期望，那么有
dp[i][j]=i/n*j/n*(dp[i][j]+1)+(n-i)/n*j/n*(dp[i+1][j]+1)+i/n*(n-j)/n*(dp[i][j+1]+1)+(n-i)/n*(n-j)/n*(dp[i+1][j+1]+1)，然后化简下得到dp[i][j]的关系式，dp[n][n]=0，从后往前推，假设此时一开始已经有a行b列满足条件，那么答案即为dp[a][b]。

代码如下

#include<iostream>#include<cstdio>#include<algorithm>  #include<cstring>using namespace std;#define LL long long #define ULL unsigned long long const int maxn=2005;int vis1[maxn],vis2[maxn];double dp[maxn][maxn];int main() {    memset(vis1,0,sizeof(vis1));    memset(vis2,0,sizeof(vis2));    memset(dp,0,sizeof(dp));    int n,m;    scanf("%d%d",&n,&m);    int a=0,b=0;    while(m--)    {        int x,y;        scanf("%d%d",&x,&y);        if(!vis1[x]) a++,vis1[x]=1;        if(!vis2[y]) b++,vis2[y]=1;     }    int N=n*n;    for(int i=n;i>=a;i--)    for(int j=n;j>=b;j--)    {        if(i==n&&j==n) continue;        dp[i][j]=(n-i)*j*(dp[i+1][j]+1)/N+i*(n-j)*(dp[i][j+1]+1)/N+(n-i)*(n-j)*(dp[i+1][j+1]+1)/N;        dp[i][j]=(dp[i][j]+i*j*1.0/N)/(1-i*j*1.0/N);    }    printf("%.5lf\n",dp[a][b]);    return 0;}