CodeForces 148D Bag of mice [概率DP]

最新推荐文章于 2021-09-03 21:27:41 发布

GrassTreeFlower

最新推荐文章于 2021-09-03 21:27:41 发布

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分类专栏：概率DP/概率/期望文章标签： ACM math codeforces DP 概率DP

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本文介绍了一种利用概率动态规划解决龙与公主新年赌局问题的方法。通过分析袋中不同颜色老鼠被抓取的概率，计算出公主赢得比赛的概率。

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Description

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.

题意：

龙与女王因是否去看精灵跳舞产生了分歧，于是决定采取从袋子里抓老鼠的方法来决定去不去看，袋子里有w只白老鼠，b只黑老鼠，谁先抓到白老鼠谁赢，而龙在从袋子里抓老鼠的时候会有一只老鼠会跑掉，现在女王先抓，问最后女王赢的概率。（一开始题意读不懂，后来才知道draw有抓的意思）

范围:

w,b<=1000

解法：

知道了具体的题意就会发现这是一道很明显的概率DP，DP[W][B]表示现在女王先手，且还有W只白老鼠，B只黑老鼠，此时女王赢的概率。

于是只需枚举女王与龙抓到的老鼠颜色即可转移，具体参考代码，注意w与b可能在转移时减为负数，需要判掉。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

double dp[1111][1111];
bool vis[1111][1111];
int n,m;
double dfs(int i,int j){
    if(i==0)return 0.0;
    if(j==0)return 1.0;
    if(vis[i][j])return dp[i][j];
    
    vis[i][j]=1;
    dp[i][j]=double(i)/double(i+j);
    if(i>=1&&j>=2){
        dp[i][j]+=double(j)/double(i+j)*double(j-1)/double(i+j-1)*
                  double(i)/double(i+j-2)*dfs(i-1,j-2);
    }
    if(j>=3){
        dp[i][j]+=double(j)/double(i+j)*double(j-1)/double(i+j-1)*
                  double(j-2)/double(i+j-2)*dfs(i,j-3);
    }


    return dp[i][j];
}
int main(){
    scanff(n);
    scanff(m);
    double ans=0;

    printf("%.10f\n",dfs(n,m));

    return 0;
}