Codeforces Round #105 D. Bag of mice probability dp,

Codeforces Round #105 D. Bag of mice probability dp,

Http://codeforces.com/contest/148/problem/D

The question is that dragon and Princess take turns pumping mice from the bag, W white teachers in the bag, and D black teachers. The Princess smokes first, and the first pulls the white mouse to win, the dragon will randomly run out of a mouse each time it is pumped. Give W and D for you to find the probability of the princess's victory.

For dp [w] [d], it indicates that there are w white mice, d black mice, and the probability of the princess's victory, if the princess pulls out the white mouse for the first time, the probability is w/(w + d). If the princess fails to win the princess, and the dragon cannot win the White Rat, rec = d/(w + d) * (D-1)/(w + D-1 ). The mouse that runs out may be white or black. If the mouse is white, then the W-1 is left white mouse D-2 only black mouse, then princess first hand, probability is rec * w/(w + D-2) * dp [W-1] [D-2]. If it is a black rat, then the remaining w white mouse D-3 only black rat, the same princess first hand, the probability is rec * (D-2)/(w + D-2) * dp [w] [D-3];

 

Finally, the dp transfer equation is obtained:

Dp [I] [j] = I/(I + j) + rec * (I/(I + J-2) * dp [I-1] [J-2] + (J-2) /(I + j-2) * dp [I] [J-3]);

Rec = j/(I + j) * (j-1)/(I + j-1 );

/*********************************************** ** problem ID    : cf_#105D.cpp ** create time    : Wed Jul 22 20:22:24 2015 ** auther name    : xuelanghu ** auther blog    : blog.csdn.net/xuelanghu407 **********************************************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int n, m;double dp[1010][1010];int main () {        cin >> n >> m;        for (int i=0; i<=n; i++) {        dp[i][0] = 1.0;    }    for (int i=0; i<=m; i++) {        dp[0][i] = 0.0;    }        for (int i=1; i<=n; i++) {        dp[i][1] = (double)i / (i+1);        for (int j=2; j<=m; j++) {            dp[i][j] = (double)i / (i+j);            double tmp = (double)j* (double)(j-1)  / (i+j) / (i+j-1);            if (j>2) dp[i][j] += tmp * (double)(j-2) / (i+j-2) * dp[i][j-3];              dp[i][j] += tmp * (double)i / (i+j-2) * dp[i-1][j-2];        }    }        printf("%.10lf\n", dp[n][m]);        return 0;}code

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