H - Grid 1

Problem Statement

There is a grid with $\mathrm{HH}$ horizontal rows and $\mathrm{WW}$ vertical columns. Let $(i,j)(i,j)$ denote the square at the $\mathrm{ii}$-th row from the top and the $\mathrm{jj}$-th column from the left.

For each $\mathrm{ii}$ and $\mathrm{jj}$ ($\mathrm{1\le i\le H1\le i\le H}$, $\mathrm{1\le j\le W1\le j\le W}$), Square $(i,j)(i,j)$ is described by a character ${\mathrm{ai,jai,j}}^{}$. If ${\mathrm{ai,jai,j}}^{}$ is ., Square $(i,j)(i,j)$ is an empty square; if ${\mathrm{ai,jai,j}}^{}$ is #, Square $(i,j)(i,j)$ is a wall square. It is guaranteed that Squares $(1,1)(1,1)$ and $(H,W)(H,W)$ are empty squares.

Taro will start from Square $(1,1)(1,1)$ and reach $(H,W)(H,W)$ by repeatedly moving right or down to an adjacent empty square.

Find the number of Taro's paths from Square $(1,1)(1,1)$ to $(H,W)(H,W)$. As the answer can be extremely large, find the count modulo ${\mathrm{109+7109+7}}^{}$.

Constraints

• $\mathrm{HH}$ and $\mathrm{WW}$ are integers.
• $\mathrm{2\le H,W\le 10002\le H,W\le 1000}$
• ${\mathrm{ai,jai,j}}^{}$ is . or #.
• Squares $(1,1)(1,1)$ and $(H,W)(H,W)$ are empty squares.

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Input

Input is given from Standard Input in the following format:

$\mathrm{HH}$ $\mathrm{WW}$
${\mathrm{a1,1a1,1}}^{}$$\dots \dots$${\mathrm{a1,Wa1,W}}^{}$
$::$
${\mathrm{aH,1aH,1}}^{}$$\dots \dots$${\mathrm{aH,WaH,W}}^{}$

Output

Print the number of Taro's paths from Square $(1,1)(1,1)$ to $(H,W)(H,W)$, modulo ${\mathrm{109+7109+7}}^{}$.

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Sample Input 1 Copy

Copy

3 4
...#
.#..
....

Sample Output 1 Copy

Copy

3

There are three paths as follows:

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Sample Input 2 Copy

Copy

5 2
..
#.
..
.#
..

Sample Output 2 Copy

Copy

0

There may be no paths.

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Sample Input 3 Copy

Copy

5 5
..#..
.....
#...#
.....
..#..

Sample Output 3 Copy

Copy

24

---

Sample Input 4 Copy

Copy

20 20
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................
....................

Sample Output 4 Copy

Copy

345263555

Be sure to print the count modulo ${\mathrm{109+7109+7}}^{}$.

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <unordered_set>
#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 1e3 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}
bool vis[N][N];
queue<pii> q;
int cnt[N][N];
int main()
{
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    for (int i = 0; i < n; i++)
        cin >> g[i];
    q.push({ 0,0 });
    vis[0][0] = 1;
    cnt[0][0] = 1;
    while (!q.empty())
    {
        pii f = q.front();
        q.pop();
        for (int i = 0; i < 2; i++)
        {
            int x = f.first + dir[i][0], y = f.second + dir[i][1];
            if (x >= 0 && x < n && y >= 0 && y < m && g[x][y] == '.')
            {
                cnt[x][y] = (cnt[f.first][f.second] + cnt[x][y]) % mod;
                if (!vis[x][y])
                {
                    vis[x][y] = 1;
                    q.push({ x,y });
                }
            }
        }
    }
    cout << cnt[n - 1][m - 1] << endl;
    return 0;
}