Bag of mice - CodeForces 148 D 概率dp-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/43818457

龙和公主通过抽签决定是否抓到白老鼠来决定胜负，公主先手，每轮龙抓后随机一只老鼠离开袋子。计算公主赢得比赛的概率。

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The dragon and the princess are arguing about what to do on the New Year's Eve. The dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they should just go to bed early. They are desperate to come to an amicable agreement, so they decide to leave this up to chance.

They take turns drawing a mouse from a bag which initially contains w white and b black mice. The person who is the first to draw a white mouse wins. After each mouse drawn by the dragon the rest of mice in the bag panic, and one of them jumps out of the bag itself (the princess draws her mice carefully and doesn't scare other mice). Princess draws first. What is the probability of the princess winning?

If there are no more mice in the bag and nobody has drawn a white mouse, the dragon wins. Mice which jump out of the bag themselves are not considered to be drawn (do not define the winner). Once a mouse has left the bag, it never returns to it. Every mouse is drawn from the bag with the same probability as every other one, and every mouse jumps out of the bag with the same probability as every other one.

Input

The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

Output

Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed10 - 9.

Sample test(s)

input
1 3

output
0.500000000

input
5 5

output
0.658730159

题意：龙和公主抓老鼠，公主先手，谁先抓到白老鼠谁赢，龙抓完后，会随机跑走一个老鼠，如果没有老鼠时，龙获胜，问公主赢的概率。

思路：dp[i][j][0|1]表示还剩i个白老鼠，j个黑老鼠时，公主|龙先手赢的概率。转移方程见代码，挺简单的。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
double dp[1010][1010][2];
bool vis[1010][1010][2];
int n,m;
double dfs(int i,int j,int f)
{
    if(vis[i][j][f])
      return dp[i][j][f];
    vis[i][j][f]=1;
    if(i<0 || j<0)
    {
        dp[i][j][f]=0;
        return 0;
    }
    if(f==0)
      dp[i][j][f]=1.0*i/(i+j)+1.0*j/(i+j)*(1-dfs(i,j-1,1));
    else
      dp[i][j][f]=1.0*i/(i+j)+1.0*j/(i+j)*(1-dfs(i-1,j-1,0)*i/(i+j-1)-dfs(i,j-2,0)*(j-1)/(i+j-1));
    return dp[i][j][f];
}
int main()
{
    int i,j,k;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        dp[i][0][0]=1;vis[i][0][0]=1;
        dp[i][0][1]=1;vis[i][0][1]=1;
    }
    for(i=0;i<=m;i++)
    {
        dp[0][i][0]=0;vis[0][i][0]=1;
        dp[0][i][1]=1;vis[0][i][1]=1;
    }
    printf("%.10f\n",dfs(n,m,0));
}