Cf148d--bag of mice+ probability expectation DP

First probability expectation DP:)

In fact, and the general DP is similar, as long as the status of the election is OK.

Definition dp[i][j] means I have only white mouse J Black Mouse when the princess won the probability.

Then: 1. Princess Choose white Mouse, direct win, probability: i/(I+J)

2. The princess chose the black Mouse

1) The Dragon chose the black Mouse, fled the black mouse; probability: j/(i+j) * (j-1)/(i+j-1) * (j-2)/(I+j-2)

2) Dragon Choose Black Mouse, Escape white mouse; probability: j/(i+j) * (j-1)/(i+j-1) *i/(i+j-2)

3) Long Mouse, so the princess will lose, do not consider

Then dp[i][j] equals the sum of the above probabilities.

When initialized, the probability of winning is 1 if only microscope

If the black mouse wins, the odds are 0.

The code is as follows:

#include <iostream> #include <cstdio> #include <cstring>using namespace std;double dp[1100][1100]; int main () {    int w,b;    while (scanf ("%d%d", &w,&b)!=eof)    {for        (int i=1;i<=w;i++)           dp[i][0]=1;        for (int j=1;j<=b;j++)           dp[0][j]=0;        for (int i=1;i<=w;i++) for          (int j=1;j<=b;j++)          {              dp[i][j]=1.0*i/(i+j);              if (j>=3)                 dp[i][j]+=j*1.0/(i+j) * (j-1) *1.0/(i+j-1) * (j-2) *1.0/(i+j-2) *dp[i][j-3];              if (j>=2)                 dp[i][j]+=j*1.0/(i+j) * (j-1) *1.0/(i+j-1) *i*1.0/(i+j-2) *dp[i-1][j-2];          }        printf ("%.9lf\n", Dp[w][b]);    }  return 0;}

Cf148d--bag of mice+ probability expectation DP