codeforces 148 D Bag of mice(概率dp)

题目大意：原来袋子里有w只白鼠和b只黑鼠龙和王妃轮流从袋子里抓老鼠。谁先抓到白色老师谁就赢。王妃每次抓一只老鼠，龙每次抓完一只老鼠之后会有一只老鼠跑出来。每次抓老鼠和跑出来的老鼠都是随机的。如果两个人都没有抓到白色老鼠则龙赢。王妃先抓。问王妃赢的概率。

解题思路：dp[i][j]表示现在轮到王妃抓时有i只白鼠，j只黑鼠，王妃赢的概率

王妃要想赢必须：第i次的时候王妃抓到白鼠，否则这个龙要抓到黑鼠王妃才有可能这轮不输，而将会跑出去的老鼠有可能是白的，也有可能是黑色的。所以就产生了不同的状态。记忆化搜一下。

D. Bag of mice

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
//#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898

const int maxn = 1010;

using namespace std;

double dp[maxn][maxn];

double dfs(int x, int y)
{
if(dp[x][y] > 0)
return dp[x][y];
if(x < 0 || y < 0)
return 0;
if(x == 0)
return 0;
if(y == 0)
return 1;
double a = x*1.0;
double b = y*1.0;
double c = a+b;
dp[x][y] = a/c;
if(x+y >= 3)
{
dp[x][y] += (dfs(x-1, y-2)*(b/(c))*((b-1)/(c-1))*(a/(c-2)) + dfs(x, y-3)*(b/(c))*((b-1)/(c-1))*((b-2)/(c-2)));
}
return dp[x][y];

}

int main()
{
int w, b;
memset(dp, 0, sizeof(dp));
while(cin >>w>>b)
{
printf("%.10lf\n",dfs(w,b));
}
return 0;
}