E. Rock Is Push
You are at the top left cell (1,1) of an n×m labyrinth. Your goal is to get to the bottom right cell (n,m). You can only move right or down, one cell per step. Moving right from a cell (x,y) takes you to the cell (x,y+1), while moving down takes you to the cell (x+1,y).
Some cells of the labyrinth contain rocks. When you move to a cell with rock, the rock is pushed to the next cell in the direction you're moving. If the next cell contains a rock, it gets pushed further, and so on.
The labyrinth is surrounded by impenetrable walls, thus any move that would put you or any rock outside of the labyrinth is illegal.
Count the number of different legal paths you can take from the start to the goal modulo 109+7. Two paths are considered different if there is at least one cell that is visited in one path, but not visited in the other.
Input
The first line contains two integers n,m — dimensions of the labyrinth (1≤n,m≤2000).
Next n lines describe the labyrinth. Each of these lines contains m characters. The j-th character of the i-th of these lines is equal to "R" if the cell (i,j) contains a rock, or "." if the cell (i,j) is empty.
It is guaranteed that the starting cell (1,1) is empty.
Output
Print a single integer — the number of different legal paths from (1,1) to (n,m) modulo 109+7.
Examples
input
1 1
.
output
1
input
2 3
...
..R
output
0
input
4 4
...R
.RR.
.RR.
R...
output
4
Note
In the first sample case we can't (and don't have to) move, hence the only path consists of a single cell (1,1).
In the second sample case the goal is blocked and is unreachable.
Illustrations for the third sample case can be found here: https://subdomain.codeforc.es/menci/assets/rounds/1225/index.html
The meaning of problems
A n * m matrix, there are a pile of boxes, you can Sokoban, continuous box you can push.
Ask you from (1,1) to (n, m) has the number of the program how many different paths.
answer
Defined:
DP [I] [J] [0] represents the (i, j) to go down to reach the end of the program number.
dp [i] [j] [ 1] represents the number of the program reaches the end away from the (i, j) to the right.
Comparative apparent
dp [i] [j] [ 0] = dp [i + 1] [j] [1] + dp [i + 2] [j] [1] + .... + dp [i + x ] [j] [1], until the (i + x + 1, j ) is pushed in the end of a box, a can not Sokoban.
Similarly dp [i] [j] [1] it is true.
Obviously this lump behind the prefix and can optimize it, and then you can become n ^ dp 2 is shifted.
Code
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
int n,m;
const int mod = 1e9+7;
char a[maxn][maxn];
// 0 for down;1 for right
int num[maxn][maxn][2],dp[maxn][maxn][2],sum[maxn][maxn][2];
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%s",a[i]+1);
}
if(n==1&&m==1&&a[1][1]=='.'){
cout<<"1"<<endl;
return 0;
}
if(a[1][1]=='R'||a[n][m]=='R'){
cout<<"0"<<endl;
return 0;
}
for(int i=n;i>=1;i--){
for(int j=m;j>=1;j--){
if(a[i][j]=='R'){
num[i][j][0]+=1;
num[i][j][1]+=1;
}
num[i][j][0]+=num[i+1][j][0];
num[i][j][1]+=num[i][j+1][1];
}
}
dp[n][m][0]=1;dp[n][m][1]=1;sum[n][m][0]=1;sum[n][m][1]=1;
for(int i=n;i>=1;i--){
for(int j=m;j>=1;j--){
if(i==n&&j==m)continue;
dp[i][j][0]=(sum[i+1][j][0]-sum[n-num[i+1][j][0]+1][j][0])%mod;
dp[i][j][1]=(sum[i][j+1][1]-sum[i][m-num[i][j+1][1]+1][1])%mod;
sum[i][j][0]=(sum[i+1][j][0]+dp[i][j][1])%mod;
sum[i][j][1]=(sum[i][j+1][1]+dp[i][j][0])%mod;
}
}
cout<<(dp[1][1][0]+dp[1][1][1]+2ll*mod)%mod<<endl;
}