挂个写时候的思路。

考虑 $L_{l,r},R_{l,r}$ 为对于 $(l,r)$ 所能延伸到最左边的 $(l,L_{l,r})$ 和最右边的 $(l,R_{l,r})$。$g_{l,r}$ 为 $(l,r)$ 所能延伸到最右上的 $(l-g_{l,r},r+g_{l,r})$。

倘若枚举 $l,r$ 为左下角，于是答案只可能在 $1 \sim \min(R_{l,r},g_{l,r})$ 中出现。而对于每个可能的答案 $i$，又需要保证 $L_{l-i,r+i} \leq l$，这个可以枚举每条斜线二位数点做。

复杂度 $O(n^2 \log n)$。

#include <iostream>
#include <cassert>
#include <queue>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <random>
#include <ctime>
#include <queue>
#include <map>
#include <set>
using namespace std;
// #define int long long
#define pii pair<int, int>
#define eb emplace_back
#define F first
#define S second
#define test(x) cout << "Test: " << (x) << '\n'
#define lowbit(x) (x & -x)
#define debug puts("qwq");
#define open(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define close fclose(stdin);fclose(stdout);
namespace FastIO {
	template <typename T = int>
	inline T read() {
		T s = 0, w = 1;
		char c = getchar();
		while (!isdigit(c)) {
			if (c == '-') w = -1;
			c = getchar();
		} 
		while (isdigit(c)) s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
		return s * w;
	}
	template <typename T>
	inline void read(T &s) {
		s = 0;
		int w = 1;
		char c = getchar();
		while (!isdigit(c)) {
			if (c == '-') w = -1;
			c = getchar();
		}
		while (isdigit(c)) s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
		s = s * w;
	}
	template <typename T, typename... Arp> inline void read(T &x, Arp &...arp) {
		read(x), read(arp...);
	}
	template <typename T>
	inline void write(T x, char ch = '\n') {
		if (x < 0) x = -x, putchar('-');
		static char stk[25];
		int top = 0;
		do {
			stk[top++] = x % 10 + '0', x /= 10;
		} while (x);
		while (top) putchar(stk[--top]);
		putchar(ch);
		return ;
	}
	template <typename T>
	inline void smax(T &x, T y) {
		if (x < y) x = y;
	}
	template <typename T>
	inline void smin(T &x, T y) {
		if (x > y) x = y;
	}
	void quit() {
		exit(0);
	}
} using namespace FastIO;
const int N = 3030;
int L[N][N], R[N][N], n, m, ans[N];
char c[N][N]; pii Q[N];
struct BIT {
	int c[N];
	void clear() { memset(c, 0, sizeof c); }
	void add(int x, int s) {
		while (x < N) {
			c[x] += s; x += lowbit(x);
		}
	}
	int ask(int x) {
		int ans = 0;
		while (x) {
			ans += c[x]; x -= lowbit(x);
		} return ans;
	}
	int query(int l, int r) { return ask(r)-ask(l-1); }
} bit;
signed main() {
	read(n, m);
	for (int i = 1; i <= n; ++i) scanf("%s", c[i]+1);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1, last = 0; j <= m; ++j) {
			if (c[i][j] == '.') last = j;
			L[i][j] = last+1;
		}
		for (int j = m, last = m+1; j; --j) {
			if (c[i][j] == '.') last = j;
			R[i][j] = last-1;
		}
	} int sum = 0;
	for (int i = 1; i <= m; ++i) {
		bit.clear();
		int x = 1, y = i, len = 0, lst = i;
		while (x <= n && y) {
			if (c[x][y] == 'z') Q[++len] = {min(lst, R[x][y]), y};
			else lst = y-1;
			x++; y--;
		} sort(Q+1,Q+1+len,greater<pii>()); Q[len+1].F=0;
		x = 1; y = i; int l=1,r;
		while (x <= n && y) {
			for (r = l; Q[r].F == y; ++r);
			for (int k = l; k < r; ++k) sum += -bit.query(1, Q[k].S);
			if (c[x][y] == 'z') { bit.add(L[x][y], 1); sum += bit.query(1, y); }
			l = r;
			x++; y--; 
		}
	}
	for (int i = 2; i <= n; ++i) {
		bit.clear();
		int x = i, y = m, len = 0, lst = i;
		while (x <= n && y) {
			if (c[x][y] == 'z') Q[++len] = {min(lst, R[x][y]), y};
			else lst = y-1;
			x++; y--;
		} sort(Q+1,Q+1+len,greater<pii>()); Q[len+1].F=0;
		x = i; y = m; int l=1,r;
		while (x <= n && y) {
			for (r = l; Q[r].F == y; ++r);
			for (int k = l; k < r; ++k) sum += -bit.query(1, Q[k].S);
			if (c[x][y] == 'z') { bit.add(L[x][y], 1); sum += bit.query(1, y); }
			l = r;
			x++; y--;
		}
	} write(sum);
	return 0;
}