标签:ch move robot cell Broken rg CF24D row
题意
D. Broken robottime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou received as a gift a very clever robot walking on a rectangular board. Unfortunately, you understood that it is broken and behaves rather strangely (randomly). The board consists of N rows and M columns of cells. The robot is initially at some cell on the i-th row and the j-th column. Then at every step the robot could go to some another cell. The aim is to go to the bottommost (N-th) row. The robot can stay at it's current cell, move to the left, move to the right, or move to the cell below the current. If the robot is in the leftmost column it cannot move to the left, and if it is in the rightmost column it cannot move to the right. At every step all possible moves are equally probable. Return the expected number of step to reach the bottommost row.
InputOn the first line you will be given two space separated integers N and M (1 ≤ N, M ≤ 1000). On the second line you will be given another two space separated integers i and j (1 ≤ i ≤ N, 1 ≤ j ≤ M) — the number of the initial row and the number of the initial column. Note that, (1, 1) is the upper left corner of the board and (N, M) is the bottom right corner.
OutputOutput the expected number of steps on a line of itself with at least 4 digits after the decimal point.
ExamplesInputCopy10 10OutputCopy
10 4
0.0000000000InputCopy
10 14OutputCopy
5 14
18.0038068653
分析
参照M_sea的博客。
设\(f[i][j]\)表示\((i,j)\)到最后一排的移动步数的期望。
容易得出:
\(f[i][1]=\frac{1}{3}(f[i][1]+f[i][2]+f[i-1][1])+1\)
\(f[i][m]=\frac{1}{3}(f[i][m]+f[i][m-1]+f[i-1][m])+1\)
\(f[i][j]=\frac{1}{4}(f[i][j]+f[i][j-1]+f[i][j+1]+f[i-1][j]\quad (1<j<m)\)
发现列与列的转移是有后效性的。
所以用有后效性DP的基本方法——DP套高斯消元。
但是发现系数矩阵中每行只有几个数要消,而且非常有规律。所以可以\(O(m)\)解。
细节见代码。注意要特判\(m=1\)的情况。
代码
第一次打特殊的高斯消元……学到了呀。
#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read(){
rg T data=0,w=1;rg char ch=getchar();
while(!isdigit(ch)) {if(ch=='-') w=-1;ch=getchar();}
while(isdigit(ch)) data=data*10+ch-'0',ch=getchar();
return data*w;
}
template<class T>il T read(rg T&x) {return x=read<T>();}
typedef long long ll;
co int N=1002;
int n,m,x,y;
double f[N][N],d[N][N];
void work(){
for(int i=1;i<=m;++i){
double w=1/d[i][i];
d[i][i]*=w,
d[i][m+1]*=w;
if(i==m) break;
d[i][i+1]*=w;
w=d[i+1][i]/d[i][i];
d[i+1][i]-=w*d[i][i],
d[i+1][i+1]-=w*d[i][i+1],
d[i+1][m+1]-=w*d[i][m+1];
}
for(int i=m-1;i;--i)
d[i][m+1]-=d[i+1][m+1]*d[i][i+1];
}
int main(){
// freopen(".in","r",stdin),freopen(".out","w",stdout);
read(n),read(m),read(x),read(y);
for(int i=n-1;i>=x;--i){
d[1][1]=d[m][m]=-2/3.0,
d[1][2]=d[m][m-1]=1/3.0;
for(int j=2;j<m;++j)
d[j][m+1]=-f[i+1][j]/4.0-1,
d[j][j]=-3/4.0,
d[j][j-1]=d[j][j+1]=1/4.0;
if(m==1) d[1][1]=-1/2.0;
d[1][m+1]=-f[i+1][1]/3.0-1;
d[m][m+1]=-f[i+1][m]/3.0-1;
if(m==1) d[m][m+1]=-f[i+1][m]/2.0-1;
work();
for(int j=1;j<=m;++j) f[i][j]=d[j][m+1];
}
printf("%lf\n",f[x][y]);
return 0;
}
标签:ch,move,robot,cell,Broken,rg,CF24D,row 来源: https://www.cnblogs.com/autoint/p/10667284.html
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。