Codeforces CF#628 Education 8 E. Zbazi in Zeydabad

E. Zbazi in Zeydabad

time limit per test

5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

A tourist wants to visit country Zeydabad for Zbazi (a local game in Zeydabad).

The country Zeydabad is a rectangular table consisting of n rows and m columns. Each cell on the country is either 'z' or '.'.

The tourist knows this country is named Zeydabad because there are lots of ''Z-pattern"s in the country. A ''Z-pattern" is a square which anti-diagonal is completely filled with 'z' and its upper and lower rows are also completely filled with 'z'. All other cells of a square can be arbitrary.

Note that a ''Z-pattern" can consist of only one cell (see the examples).

So he wants to count the number of ''Z-pattern"s in the country (a necessary skill for Zbazi).

Now your task is to help tourist with counting number of ''Z-pattern"s.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to usegets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead ofScanner/System.out in Java.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 3000) — the number of rows and columns respectively.

Each of the next n lines contains m characters 'z' or '.' — the description of Zeydabad.

Output

Print the only integer a — the number of ''Z-pattern"s in Zeydabad.

Examples

input

4 4
zzzz
zzz.
.z..
zzzz

output

16

input

1 4
z.z.

output

2

input

2 2
zz
zz

output

5

题意：
    给出一个n*n的矩阵，问有多少个子矩阵满足副对角线、最上面的行、最下面的行都是由z构成（其他位置无关）（理解一下，这样的图形构成了一个Z的图案）。

　　

题解：
    如果预处理出每个点向左向右连续的z能够有多长，
    那么先枚举副对角线，
    由于答案肯定是由副对角线上连续的一段z贡献的，
    所以我们可以一段段z来做。
    问题转换成了，在这一段z上，有多少个点对（i，j）满足
    left[i]>=j - i, right[j] >= j - i，
    也就是说，i、j必须能够分别用left[i]、right[j]的长度分别覆盖对方。

    那么我们可以枚举j，确定有多少i对它进行了贡献。
    那么对于每个i，我们都能够知道它最远能够贡献到哪里，
    每当我们的j超过了这个i最远能贡献的距离，就将它排去，
    因为i不可能再对能够覆盖它的j左贡县。
    对于每个j，现在所有能够留下来的i都是能够覆盖它的。
    能够对它有贡献的i就是那些它能覆盖的i。

    所以保持i能覆盖当前的j这一步可以使用优先队列完成，
    统计有多少i贡献j能够用树状数组完成。

    挺水的。

　　

 #include <cstdio>
 #include <queue>
 #include <cstring>
 #include <string>
 #include <iostream>
 #include <algorithm>
 #include <cstdlib>
 using namespace std;
 #define clr(x, y) memset(x, y, sizeof(x))
 #define mk make_pair

 const int N = ;
 typedef long long ll;
 int n, m;
 char graph[N][N];
 int lef[N][N], rig[N][N];

 ll sum[N];
 priority_queue<pair<int, int> > outQue;

 int lowbit(int x) {
     return x & (-x);
 }

 int len;
 void add(ll *arr, int pos, ll val) {
     if(!pos) return;
     for(int x = pos; x <= len; x += lowbit(x)) arr[x] += val;
 }

 ll query(ll *arr, int pos) {
     ll ret = ;
     for(int x = pos; x; x -= lowbit(x)) ret += arr[x];
     return ret;
 }

 ll query(ll *arr, int lef, int rig) {
     ll b = query(arr, rig);
     ll a = query(arr, lef - );
     return b - a;
 }

 void solve() {
     for(int i = ; i <= n; ++i) {
         lef[i][] = ;
         for(int j = ; j <= m; ++j)
             if(graph[i][j] == '.') lef[i][j] = ;
             else lef[i][j] = lef[i][j - ] + ;
         lef[i][m + ] = ;
         for(int j = m; j >= ; --j)
             if(graph[i][j] == '.') rig[i][j] = ;
             else rig[i][j] = rig[i][j + ] + ;
     }

     ll ans = ;
     for(int diagon = ; diagon <= n + m - ; ++diagon) {
         int stx = max(, diagon - m + ), sty = min(diagon, m);
         int tot = ;
         bool first = true;
         for(int x = stx, y = sty; x <= n && y >= ; ++x, --y)
             if(graph[x][y] == '.') {
                 for(int i = ; i <= tot; ++i) sum[i] = ;
                 while(!outQue.empty()) outQue.pop();
                 tot = , first = true;
             } else {
                 if(first) {
                     len = , first = false;
                     for(int _x = x, _y = y; _x <= n && _y >=  && graph[_x][_y] == 'z'; ++_x, --_y)
                         ++len;
                 }
                 ++tot;
                 add(sum, tot, );
                 outQue.push(mk(-(tot + lef[x][y] - ), tot));

                 while(-outQue.top().first < tot) {
                     add(sum, outQue.top().second, -);
                     outQue.pop();
                 }

                 ll tmp = query(sum, max(, tot - rig[x][y] + ), tot);
                 ans += tmp;
             }
         for(int i = ; i <= tot; ++i) sum[i] = ;
         while(!outQue.empty()) outQue.pop();
     }

     cout << ans << endl;
 }

 int main() {
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; ++i) scanf("%s", graph[i] + );
     solve();
     return ;
 }