codeforces 148 D. Bag of mice

Posted by 111qqz on Wednesday, February 3, 2016

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``````1. 公主抽到白老鼠（之后龙不必再抽）  胜率为i/(i+j)*1
2. 公主抽到黑老鼠，龙抽到黑老鼠，跳出一只黑老鼠，胜率为j/(i+j) * (j-1)/(i+j-1) * (j-2)/(i+j-2) * f[i][j-3] (j>=3)
3. 公主抽到黑老鼠，龙抽到黑老鼠，跳出一只白老鼠，胜率为j/(i+j) * (j-1)/(i+j-1) * (i/(i+j-2) * f[i-1][j-2] (j>=2)
4. 龙抽到白老鼠，胜率为0

/* ***********************************************
Author :111qqz
Created Time :2016年02月04日 星期四 02时44分23秒
File Name :code/cf/problem/148D.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E3+7;
int w,b;
double dp[N][N];
int main()
{
#ifndef  ONLINE_JUDGE
freopen("code/in.txt","r",stdin);
#endif
cin>>w>>b;

ms(dp,0);
//dp[i][j]表示有i个白球,j个黑球的时候公主获胜的概率。
for ( int i = 1 ; i <= w ; i++) dp[i][0] = 1;
for ( int j = 1 ; j <= b ; j++) dp[0][j]= 0 ;

for ( int i = 1 ; i <= w ; i ++)
{
for ( int j = 1 ; j <= b ; j++)
{
double x = i*1.0;
double y = j*1.0;

dp[i][j]+=x/(x+y);
if (j>=3) dp[i][j] +=y/(x+y)*(y-1)/(x+y-1)*(y-2)/(x+y-2)*dp[i][j-3];

if (j>=2) dp[i][j] +=y/(x+y)*(y-1)/(x+y-1)*(x)/(x+y-2)*dp[i-1][j-2];
}
}
printf("%.10f\n",dp[w][b]);

#ifndef ONLINE_JUDGE
fclose(stdin);
#endif
return 0;
}
``````

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