Codeforces 148D Bag of mice 概率dp

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.

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袋子里有w个白鼠，b个黑鼠。第一个取出白鼠的赢。龙每次取出一个老鼠就会从袋子里随机吓出一只老鼠。

问公主赢的概率。

f[i][j][0]表示袋子里省i个白鼠j和黑鼠轮到公主取时公主的胜率。

轮到公主时，公主有两种方法获胜：

1、取出白鼠，概率为i/(i+j)。

2、取出黑鼠，概率为 j/(i+j)* f[i][j-1][1]。

轮到龙时，龙有两种方法不获胜：

1、取出黑鼠，吓跑一只白鼠，概率为 j/(i+j)【取出黑鼠】*i*(i+j-1)【吓跑白鼠】*f[i-1][j-1][0]。

2、取出黑鼠，吓跑一只黑鼠，概率为 j/(i+j)【取出黑鼠】*(j-1)*(i+j-1)【吓跑黑鼠】*f[i][j-2][0];

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-------果然还是记忆化搜索比较好理解--------------

double dfs(int w,int v,int turn)

直接搜好了,有点博弈的感觉。

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直接dp

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

double f[1111][1111][2];//左白，右黑
int w,b;

int main()
{
    scanf("%d%d",&w,&b);
    memset(f,0,sizeof(f));
    for (int i=0;i<=w;i++)
    {
        for (int j=0;j<=b;j++)
        {
            if (i!=0) f[i][j][0]= 1.0*i/(i+j);
            if (j!=0) f[i][j][0]+= 1.0*j/(i+j)*f[i][j-1][1];
            if (i>=1&&j>=1) f[i][j][1]= 1.0*j/(i+j)*1.0*i/(i+j-1)*f[i-1][j-1][0];
            if (j>=2) f[i][j][1]+= 1.0*j/(i+j)*1.0*(j-1)/(i+j-1)*f[i][j-2][0];
        }
    }
    printf("%0.9lf\n",f[w][b][0]);
    return 0;
}

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记忆化搜索

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

double f[1111][1111][2];//左白，右黑
bool vis[1111][1111][2];

double dfs(int w,int b,int turn)
{
    if (vis[w][b][turn]) return f[w][b][turn];
    double ret=0;
    if (turn==0)
    {
        if (w>=1) ret+=1.0*w/(w+b);
        if (b>=1) ret+=1.0*b/(w+b)*dfs(w,b-1,1);
    }
    else if (turn==1)
    {
        if (w>=1&&b>=1) ret+=1.0*b/(w+b)*1.0*w/(w+b-1)*dfs(w-1,b-1,0);
        if (b>=2) ret+=1.0*b/(w+b)*1.0*(b-1)/(w+b-1)*dfs(w,b-2,0);
    }
    vis[w][b][turn]=true;
    f[w][b][turn]=ret;
    return ret;
}

int main()
{
    int w,b;
    scanf("%d%d",&w,&b);
    memset(f,0,sizeof(f));
    memset(vis,0,sizeof(vis));
    printf("%0.9lf\n",dfs(w,b,0));
    return 0;
}