!codeforces 399D Painting The Wall-概率dp
2015-07-24 15:57
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题意:有一个n*n的墙,现在小明来刷墙,如果每一行每一列都至少有一个格子刷过了就停止工作,否则每次随机选一个格子,如果刷过了就不刷如果没刷过就刷,然后休息一分钟,求停止工作时时间的数学期望(开始之前已经有m个格子刷过了)
分析:概率dp
状态:dp[i][j]表示还有i行j列没刷,则它能转移到的状态是dp[i][j],dp[i-1][j-1],dp[i][j-1],dp[i-1][j-1]
转移:dp[i][j]=dp[i][j]*(n-i)(n-j)/n^2+dp[i-1][j]*(i*(n-j))/n^2+dp[i][j-1]*((n-i)*j)/n^2+dp[i-1][j-1]*(i*j)/n^2
初始化还没太懂
代码:
分析:概率dp
状态:dp[i][j]表示还有i行j列没刷,则它能转移到的状态是dp[i][j],dp[i-1][j-1],dp[i][j-1],dp[i-1][j-1]
转移:dp[i][j]=dp[i][j]*(n-i)(n-j)/n^2+dp[i-1][j]*(i*(n-j))/n^2+dp[i][j-1]*((n-i)*j)/n^2+dp[i-1][j-1]*(i*j)/n^2
初始化还没太懂
代码:
#include<iostream> #include<cstdio> using namespace std; double dp[2010][2010]; int n,m,a[2010],b[2010]; int main() { cin>>n>>m; int x,y; int l=n,r=n; for(int i=0;i<m;i++){ cin>>x>>y; if(!a[x]) l--; if(!b[y]) r--; a[x]=1,b[y]=1; } for(int i=1;i<=l;i++) dp[i][0]=dp[i-1][0]+(double)n/i; for(int j=1;j<=r;j++) dp[0][j]=dp[0][j-1]+(double)n/j; for(int i=1;i<=l;i++){ for(int j=1;j<=r;j++){ dp[i][j]=(dp[i-1][j]*i*(n-j)+n*n+dp[i][j-1]*j*(n-i)+dp[i-1][j-1]*i*j)/(n*n-(n-i)*(n-j)); } } printf("%0.8lf\n",dp[l][r]); }
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