CodeForces - 148D Bag of mice（动态规划）

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

Note
Let's go through the first sample. The probability of the princess drawing a white mouse on her first turn and winning right away is 1/4. The probability of the dragon drawing a
black mouse and not winning on his first turn is 3/4 * 2/3 = 1/2. After this there are two mice left in the bag — one black and one white; one of them jumps out, and the other is drawn by the princess on her second turn. If the princess' mouse is white, she
wins (probability is 1/2 * 1/2 = 1/4), otherwise nobody gets the white mouse, so according to the rule the dragon wins.

分析是摘抄的：/article/2634425.html

dp[x][y]:现在有x个白老鼠，y个黑老鼠，公主赢的概率。
那么：
如果公主直接拿到白老鼠，概率为x/(x+y),公主赢。
如果公主拿到黑老鼠，概率为y/(x+y),那么公主如果想赢，龙必须拿到黑老鼠，概率为(y-1)/(x+y-1);
那么逃跑的老鼠为黑老鼠的概率为(y-2)/(x+y-2),为白老鼠的概率为(x)/(x+y-2);
那么dp[x][y]=x/(x+y)+y/(x+y) * (y-1)/(x+y-1) * ( (y-2)/(x+y-2) * dp[x][y-3] + (x)/(x+y-2) * dp[x-1][y-2] );
记忆化深搜也行，直接递推DP也行。

#include<iostream>
#include<cstring>
#include<string>
#include<iomanip>
using namespace std;
const int MAXN = 1100;
double dp[MAXN][MAXN];
double dfs(int x, int y)
{
	if (dp[x][y] > -0.5) return dp[x][y];
	if (x < 0 || y<0) return 0;
	if (x == 0) return 0;
	if (y == 0) return 1;	
	double w, b;
	w = x*1.0;
	b = y*1.0;
	dp[x][y] = w / (w + b);
	if (x + y >= 3)
	{
		dp[x][y] += b / (w + b)*(b - 1) / (w + b - 1)*(dfs(x-1, y - 2)* w / (w + b - 2) + dfs(x, y - 3)*(b - 2) / (w + b - 2));
	}
	return dp[x][y];
}

int main()
{
	int w, b;
	while (cin >> w >> b)
	{
		memset(dp, -1, sizeof(dp));
		cout <<fixed<<setprecision(10)<< dfs(w, b) << endl;
	}
}