Codeforces Round #105 (Div. 2) D. Bag of mice 概率dp

题目链接：

http://codeforces.com/problemset/problem/148/D

D. Bag of mice

time limit per test2 seconds
memory limit per test256 megabytes
#### 问题描述
> 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.

输入

The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

输出

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.

样例输入

1 3

样例输出

0.500000000

题意

公主和龙博弈，一个箱子里面有w只白色的老鼠，b只黑色的老鼠，每一轮可以向箱子里面随机抓一只老鼠，如果抓出来的是白鼠，则直接胜出，否则换人，龙抓的时候会同时下跑一只箱子里面的老鼠，这只老鼠也是随机的一只，公主比较温柔，不会下走额外的老鼠。公主先开始抓，问公主胜出的概率。

题解

dp[i][j][0]:表示现在是公主抓，并且箱子里面还剩i只白鼠、j只黑鼠的概率。

dp[i][j][1]:表示现在是龙抓，并且箱子里面还剩i只白鼠、j只黑鼠的概率。

代码

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1111;
int n;
int w,b;
double dp[maxn][maxn][2];

void init(){
    clr(dp,0);
}

int main() {
    init();
    scf("%d%d",&w,&b);
    dp[w][b][1]=0;
    dp[w][b][0]=1;
    double ans=0.0;
    for(int i=w;i>=0;i--){
        for(int j=b;j>=0;j--){
            ///必输
            if(i==0){ continue; }

            ///这局分不出胜负
            if(j-1>=0) dp[i][j-1][1]+=j*1.0/(i+j)*dp[i][j][0];

            ///这局就赢的概率
            ans+=i*1.0/(i+j)*dp[i][j][0];

            ///抓到一只黑的，同时跑了一只黑的
            if(j-2>=0) dp[i][j-2][0]+=1.0*j*(j-1)/((i+j)*(i+j-1))*dp[i][j][1];

            ///抓到一只黑的，同时跑了一只白的
            if(j-1>=0) dp[i-1][j-1][0]+=1.0*i*j/((i+j)*(i+j-1))*dp[i][j][1];

        }
    }

    prf("%.9lf\n",ans);

    return 0;
}

//end-----------------------------------------------------------------------

记忆化搜索：

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef int LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1111;

int w,b;
double dp[maxn][maxn][2];
double dfs(int x,int y,int z){
    if(x<=0||y<0) return 0;
    if(dp[x][y][z]>=eps) return dp[x][y][z];

    double& res=dp[x][y][z]=0;
    if(z==0){
        res+=1.0*x/(x+y);
        if(y>0) res+=1.0*y/(x+y)*dfs(x,y-1,z^1);
    }else{
        if(y>1) res+=1.0*y/(x+y)*(y-1)/(x+y-1)*dfs(x,y-2,z^1);
        if(y>0) res+=1.0*y/(x+y)*x/(x+y-1)*dfs(x-1,y-1,z^1);
    }
    return res;
}

int main() {
    clr(dp,-1);
    scf("%d%d",&w,&b);
    prf("%.9lf\n",dfs(w,b,0));
    return 0;
}

//end-----------------------------------------------------------------------