Codeforces-148D Bag of mice （概率DP）

D. Bag of mice

http://www.codeforces.com/problemset/problem/148/D

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.

Examples

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.

题目大意：有一个装有w只白鼠和b只黑鼠的袋子，先抓出一只白鼠的人赢，若最后没有老鼠且没人赢，则恶龙赢，求公主先抓获胜的概率？

解法一：概率DP

大致思路：设p[i][j]表示公主在有i只白鼠，j只黑鼠时先手获胜概率，p[i][j]表示恶龙在有i只白鼠，j只黑鼠时先手获胜概率

则状态转移方程为：d[i][j]=i/(i+j)+j/(i+j)*(1-(i/(i+j-1)*p[i-1][j-1]+(j-1)/(i+j-1)*p[i][j-2]));

                                 p[i][j]=i/(i+j)+j/(i+j)*(1-d[i][j-1]);

即：d[i][j]=1-(i*j*p[i-1][j-1]+j*(j-1)*p[i][j-2])/((i+j)*(i+j-1));

       p[i][j]=1-j*d[i][j-1]/(i+j);

边界条件为：d[i][0]=1;    d[0][j]=1;

                   p[i][0]=1;    p[0][j]=0;

而且：黑鼠只有1只时得特殊处理，防止越界

#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

const int MAXN=1005;
const double EPS=0.000001;

int w,b;
double p[MAXN][MAXN];//p[i][j]表示公主在有i只白鼠，j只黑鼠时先手获胜概率
double d[MAXN][MAXN];//d[i][j]表示恶龙在有i只白鼠，j只黑鼠时先手获胜概率

int main() {
    while(scanf("%d%d",&w,&b)==2) {
        p[0][0]=0;
        d[0][0]=1;
        for(int i=1;i<=w;++i) {
            d[i][0]=1;
            p[i][0]=1;
            d[i][1]=1.0-p[i-1][0]/(i+1);
            p[i][1]=1.0-1.0*d[i][0]/(i+1);
        }
        for(int j=1;j<=b;++j) {
           d[0][j]=1;
           p[0][j]=0;
        }
        for(int i=1;i<=w;++i) {
            for(int j=2;j<=b;++j) {
                d[i][j]=1.0-1.0*(i*j*p[i-1][j-1]+j*(j-1)*p[i][j-2])/((i+j)*(i+j-1));
                p[i][j]=1.0-1.0*j*d[i][j-1]/(i+j);
            }
        }
        printf("%.9lf\n",p[w][b]);
    }
    return 0;
}

解法二：概率DP

看了题解后发现自己离熟练DP还差很远，其实只用记住公主获胜的概率即可，浪费了一倍的空间

设dp[i][j]表示公主在有i只白鼠，j只黑鼠时先手获胜概率

则dp[i][j]可以转移至一下四种状态：

①公主抓住一只白鼠，公主获胜，概率为：i/(i+j)，获胜概率为：i/(i+j);

②公主抓住一只黑鼠，恶龙抓住一只白鼠，公主输，概率为：j/(i+j)*i/(i+j-1)，获胜概率为：0;

③公主抓住一只黑鼠，恶龙抓住一只黑鼠，跑出来一只白鼠，概率为：j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)，获胜概率为：j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*dp[i-1][j-2];

④公主抓住一只黑鼠，恶龙抓住一只黑鼠，跑出来一只黑鼠，概率为：j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)，获胜概率为：j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*dp[i][j-3];

边界条件为：dp[i][0]=1;    dp[0][j]=0;

#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

const int MAXN=1005;
const double EPS=0.000001;

int w,b;
double dp[MAXN][MAXN];//dp[i][j]表示公主在有i只白鼠，j只黑鼠时先手获胜概率

int main() {
    while(scanf("%d%d",&w,&b)==2) {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=w;++i) {
            dp[i][0]=1;
        }
        for(int i=1;i<=w;++i) {
            for(int j=1;j<=b;++j) {
                dp[i][j]+=1.0*i/(i+j);//公主抓住一只白鼠
                if(j>=2) {//公主抓住一只黑鼠，恶龙抓住一只黑鼠，跑出来一只白鼠
                    dp[i][j]+=1.0*j/(i+j)*(j-1)/(i+j-1)*i/(i+j-2)*dp[i-1][j-2];
                }
                if(j>=3) {//公主抓住一只黑鼠，恶龙抓住一只黑鼠，跑出来一只黑鼠
                    dp[i][j]+=1.0*j/(i+j)*(j-1)/(i+j-1)*(j-2)/(i+j-2)*dp[i][j-3];
                }
            }
        }
        printf("%.9lf\n",dp[w][b]);
    }
    return 0;
}