文章标题 coderforces 148D ： Bag of mice （概率DP）

Bag of mice

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 exceed 10 - 9.
Example
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.
题意 ： 有一个王妃和一只龙在一个袋子里抓老鼠，袋子里有w只白色老鼠和b只黑色老鼠，当谁先抓出一只白色的老鼠谁就赢，但由于老鼠害怕龙，所以龙抓完老鼠后会再随机跑出一只老鼠出来，当没有老鼠抓时是龙赢了。
分析：用dp[i,j]来表示还剩下i只白鼠，j只黑鼠，然后轮到王妃抓老鼠，王妃能取胜的概率，所以可以得到dp[i][0]是为1，因为剩下的都是白鼠，所以肯定赢，dp[0][j]为0，只剩下黑鼠，所以肯定为输，然后可以得到一下几种状态
1、直接就抓出白鼠，概率为i/(i+j）
2、抓出黑鼠，然后龙抓出白鼠，王妃输
3、抓出黑鼠，然后龙再抓出黑鼠，跑出来黑鼠，王妃赢，这个情况的前提是还剩下至少3只黑鼠
4抓出黑鼠，然后龙再抓出黑鼠，跑出来白鼠，王妃赢，这个情况的前提是还剩下至少2只白鼠。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;double dp[1005][1005];//dp[i][j]表示还剩下i只白鼠，j只黑鼠，下一把王妃抓，王妃赢的概率 int w,b; int main (){    while (cin>>w>>b){        memset (dp,0,sizeof (dp));        for (int i=1;i<=w;i++){            dp[i][0]=1;//没黑鼠，肯定赢        }        for (int i=0;i<=b;i++){            dp[0][i]=0;//没白鼠，肯定输        }        for (int i=1;i<=w;i++){            for (int j=1;j<=b;j++){                dp[i][j]+=(double)i/(i+j);//一上来直接抓到白的                 if (j>=3){//抓到黑的，对方再抓到黑的，跑出来黑的                    dp[i][j]+=(double)j/(i+j)*(double)(j-1)/(j-1+i)*(double)(j-2)/(j-2+i)*dp[i][j-3];                }                if (j>=2){//抓到黑的，对方再抓到黑的，跑出来白的                    dp[i][j]+=(double)j/(i+j)*(double)(j-1)/(j-1+i)*(double)(i)/(j-2+i)*dp[i-1][j-2];                }            }        }        printf ("%.9f\n",dp[w][b]);    }       return 0;}