CF 148D. Bag of mice (可能性DP)

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.

採用了搜索记忆化的思想，每次的问题向下归结为较小的子问题求解。

注意一下精度问题，假设精度已经达到要求，就不用再向下进行求解了。

dp[w][b]=p1+p2*tmp;

p1=w/(w+b) //公主直接赢的概率

p2=b/(w+b), b--; p*=(b/(w+b)); b--; 经过一轮，两个选手均未摸到白老鼠

tmp=dfs(w-1,b)*(w/(w+b))+dfs(w,b-1)*(b/(w+b));  //向下深搜乘以对应的概率，某种颜色老鼠吓跑的概率。。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1005
#define LL __int64
const int inf=0x1f1f1f1f;
const double eps=1e-10;
double f[N][N];
double dfs(int w,int b)  //Princess赢的概率
{
    if(w<=0)
        return 0;
    if(b<=0)
        return 1;
    if(fabs(f[w][b]+1.0)>eps)
        return f[w][b];
    double p1,p2,tmp=1;
    p1=w*1.0/(w+b); //直接赢
    p2=b*1.0/(w+b); //还没输
    b--;
    p2=p2*(b*1.0/(w+b)); //dragon 也没赢
    b--;
    if(p2>eps)       //精度不够则继续向下深搜!!
        tmp=(w*1.0/(w+b))*dfs(w-1,b)+(b*1.0/(w+b))*dfs(w,b-1);
    //printf("%.9f %.9f %.9f\n",p1,p2,tmp);
    return f[w][b+2]=p1+p2*tmp;
}
int main()
{
    int i,w,b;
    memset(f,-1,sizeof(f));
    while(scanf("%d%d",&w,&b)!=-1)
    {
        printf("%.9f\n",dfs(w,b));
    }
    return 0;
}

非递归形式的：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1005
#define LL __int64
const int inf=0x1f1f1f1f;
const double eps=1e-10;
double dp[N][N]; //dp[w][b] 代表当前情况下Princess赢的概率
void inti()
{
    memset(dp,-1,sizeof(dp));
    dp[0][0]=0;
    for(int i=1;i<N;i++)
    {
        dp[i][0]=1;
        dp[0][i]=0;
    }
}
int main()
{
    int i,j,w,b;
    double tmp;
    inti();
    while(scanf("%d%d",&w,&b)!=-1)
    {
        if(fabs(dp[w][b]+1.0)>eps)
        {
            printf("%.9f\n",dp[w][b]);
            continue;
        }
        for(i=1;i<=w;i++)
        {
            for(j=1;j<=b;j++)
            {
                dp[i][j]=i*1.0/(i+j);
                tmp=j*1.0/(i+j)*(j-1)/(i+j-1);
                if(j>=2&&i+j>3)
                    dp[i][j]+=tmp*dp[i-1][j-2]*i/(i+j-2);
                if(j>=3&&i+j>3)
                    dp[i][j]+=tmp*dp[i][j-3]*(j-2)/(i+j-2);
            }
        }
        printf("%.9f\n",dp[w][b]);
    }
    return 0;
}

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