codeforces-148D-Bag of mice-概率DP

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dp[x][y]:现在有x个白老鼠，y个黑老鼠，公主赢的概率。

那么：

如果公主直接拿到白老鼠，概率为x/(x+y),公主赢。

如果公主拿到黑老鼠，概率为y/(x+y),那么公主如果想赢，龙必须拿到黑老鼠，概率为(y-1)/(x+y-1);

那么逃跑的老鼠为黑老鼠的概率为(y-2)/(x+y-2),为白老鼠的概率为(x)/(x+y-2);

那么dp[x][y]=x/(x+y)+y/(x+y) * (y-1)/(x+y-1) * ( (y-2)/(x+y-2) * dp[x][y-3] + (x)/(x+y-2) * dp[x-1][y-2] );

记忆化深搜也行，直接递推DP也行。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stdlib.h>#include<vector>#include<cmath>#include<queue>#include<set>#include<stack>using namespace std;#define N 100010#define LL long long#define INF 0xfffffff#define maxn 1100double dp[maxn][maxn];double dos(int x,int y){    if(x<0||y<0)return 0;    if(x==0)return 0;    if(y==0)return 1;    if(dp[x][y]>-0.5)return dp[x][y];    double a,b;    a=1.0*x;    b=1.0*y;    dp[x][y]=a/(a+b);    if(x+y>=3)    {        dp[x][y]+=(b*(b-1)/((a+b)*(a+b-1)))*        (dos(x-1,y-2)*a/(a+b-2)+dos(x,y-3)*(b-2)/(a+b-2));    }  //  cout<<x<<" "<<y<<" "<<dp[x][y]<<endl;    return dp[x][y];}int main(){    int a,b;    while(~scanf("%d%d",&a,&b))    {        memset(dp,-1,sizeof(dp));        printf("%.10lf\n",dos(a,b));    }    return 0;}

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