CoderForce 148D-Bag of mice (概率DP求概率)
题目大意:美女与野兽在玩画鸽子的游戏。鸽子在用黑布遮住的笼子里,白色的有w只,黑色的有b只,每次拿出一只作画,谁先画到白色的鸽子谁就赢。美女首先画,因为野兽太丑,它每次画的时候都会吓跑一只鸽子,所有出笼子的鸽子都不在进去。求美女赢得概率。(设定假如没有人画到白色鸽子,算野兽赢)。
题目分析:这道题不难,很显然的概率DP。这是我第一次写概率DP,纪念一下。。。
代码如下:
# include<iostream> # include<cstdio> # include<vector> # include<list> # include<queue> # include<cstring> # include<set> # include<map> # include<string> # include<cmath> # include<algorithm> using namespace std; double dp[1005][1005]; double DP(int w,int b) { if(dp[w][b]!=-1.0) return dp[w][b]; if(w<=0) return dp[w][b]=0.0; if(b<=0) return dp[w][b]=1.0; double res1=(double)w/(double)(w+b); double res2=((double)b/(double)(w+b))*((double)(b-1)/(double)(w+b-1)); double t=0.0; if(b-2>0) t+=((double)(b-2)/(double)(w+b-2))*DP(w,b-3); if(w-1>0) t+=((double)(w)/(double)(w+b-2))*DP(w-1,b-2); res2*=t; return dp[w][b]=res1+res2; } int main() { int w,b; while(~scanf("%d%d",&w,&b)) { for(int i=0;i<=w;++i) for(int j=0;j<=b;++j) dp[i][j]=-1.0; printf("%.9lf\n",DP(w,b)); } return 0; }