Problem ACodeForces 148D 概率dp

Problem ACodeForces 148D 概率dp

題意：袋子裡有w只白鼠和b只黑鼠。龍和公主輪流從袋子裡抓老鼠。誰先抓到白色老師誰就贏。公主每次抓一只老鼠，龍每次抓完一只老鼠之後會有一只老鼠跑出來。每次抓老鼠和跑出來的老鼠都是隨機的。如果兩個人都沒有抓到白色老鼠則龍贏。公主先抓。問公主贏的概率。

做了這麼多概率dp的題目了，本來接的差不多了，結果一做還是不會。。。。。。

下面是看了別人的思路

win[i][j] = i * 1.0 / (i + j); //i只白老鼠j只黑老鼠時公主選白老鼠
win[i][j] += lost[i][j-1] * j * 1.0 / (i + j); //i只白老鼠j只黑老鼠時公主選黑老鼠，但公主選完黑老鼠後龍還是輸了
lost[i][j] = j * 1.0 / (i + j) * win[i-1][j-1] * (i * 1.0 / (i + j - 1)); //i只白老鼠j只黑老鼠時龍選黑老鼠，選完後跳出去只白老鼠
lost[i][j] += j * 1.0 / (i + j) * win[i][j-2] * ((j - 1) * 1.0 / (i + j - 1)); //i只白老鼠j只黑老鼠時龍選黑老鼠，選完後跳出去只黑老鼠

Description

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 Input

1 3

0.500000000

5 5

0.658730159

Hint

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.

#include
#include
#include
#include
using namespace std;
double win[1100][1100],lost[1100][1100];
int main()
{
    int w,b;
    while(~scanf("%d %d",&w,&b))
    {
    memset(win,0,sizeof(win));
    memset(lost,0,sizeof(lost));
    for (int i = 1; i <= w; ++i)
            win[i][0] = 1.0;
    for(int i=1; i<=w; i++)
        for(int j=1; j<=b; j++)
        {
           win[i][j] = i * 1.0 / (i + j) + lost[i][j-1] * j * 1.0 / (i + j);//這個是他本次贏或者本次不贏但是下次龍輸到最後一定贏
          lost[i][j]=j*1.0/(i+j)*win[i-1][j-1]*i*1.0/(i+j-1);//龍本次輸了肯定是選了黑色的，所以要用下次公主贏得乘上對應跑白色或者黑色的概率
          lost[i][j]+=j*1.0/(i+j)*win[i][j-2]*(j-1)*1.0/(i+j-1);
        }

        printf("%.9lf\n",win[w][b]);
    }
    return 0;
}