Bag of mice

Subject:
https://www.luogu.com.cn/problem/CF148D

Nn in the bag$n$ white mice and$m$ black rats,$A$ and$B\; takes$ turns from the bag, whoever catches the white first wins. $A$ randomly grabs one at a time,$After\; B$ randomly catches one, another random mouse will run out. If neither person catches white, then$B$赢. $A$ catch first, askAAThe probability of$A$ winning.

Idea:
triples $(A,i,j)$ means$i$ white mouse,$j$ black mice,$A$ first catch this state,
use$f(A,i,j)$ means$(A,i,j)$ indicates the state to start the game,AAThe probability of$A$ winning, the answer is$f(A,n,m)$。
$$\begin{array}{rl}& (A,i,j)\frac{i}{i+j}\to Awin\\ & (A,i,j)\frac{j}{i+j}\to (B,i,j-1)\end{array}$$
Represents $(A,i,j)$有$\frac{i}{i+j}$The probability of entering directly into the winning state, $\frac{j}{i+j}$The probability of entering $(B,i,j-1)$ This state. So
$$f(A,i,j)=\frac{i}{i+j}\ast 1+\frac{j}{i+j}\ast f(B,i,j-1)$$
Similarity analysis$(B,i,j)$ .
You can directly memorize the search, pay attention to the analysis of the initial value.

#include<bits/stdc++.h>
using namespace std;
const int N=1009;
int n,m;
double dp[2][N][N];
double dfs(int x,int y,int z)
{
    
    
    if(!y)
        return 0;
    if(!z&&x)
        return 1;
    if(!z)
        return 0;
    if(z<0&&x)
        return 0;
    if(dp[x][y][z]!=0)
        return dp[x][y][z];
    if(x)
        dp[x][y][z]+=(double)z/(y+z)*dfs(0,y,z-1)+(double)y/(y+z);
    else
        dp[x][y][z]+=(double)z/(y+z)*(((double)z-1)/(y+z-1)*dfs(1,y,z-2)+(double)y/(y+z-1)*dfs(1,y-1,z-1));
    return dp[x][y][z];
}
int main()
{
    
    
    scanf("%d%d",&n,&m);
    printf("%.9lf",dfs(1,n,m));
    return 0;
}