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 = 3010;
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 = 0;
    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 - 1);
    return b - a;
}

void solve() {
    for(int i = 1; i <= n; ++i) {
        lef[i][0] = 0;
        for(int j = 1; j <= m; ++j)
            if(graph[i][j] == '.') lef[i][j] = 0;
            else lef[i][j] = lef[i][j - 1] + 1;
        lef[i][m + 1] = 0;
        for(int j = m; j >= 1; --j)
            if(graph[i][j] == '.') rig[i][j] = 0;
            else rig[i][j] = rig[i][j + 1] + 1;
    }

    ll ans = 0;
    for(int diagon = 1; diagon <= n + m - 1; ++diagon) {
        int stx = max(1, diagon - m + 1), sty = min(diagon, m);
        int tot = 0;
        bool first = true;
        for(int x = stx, y = sty; x <= n && y >= 1; ++x, --y)
            if(graph[x][y] == '.') {
                for(int i = 1; i <= tot; ++i) sum[i] = 0;
                while(!outQue.empty()) outQue.pop();
                tot = 0, first = true;
            } else {
                if(first) {
                    len = 0, first = false;
                    for(int _x = x, _y = y; _x <= n && _y >= 1 && graph[_x][_y] == 'z'; ++_x, --_y)
                        ++len;
                }
                ++tot;
                add(sum, tot, 1);
                outQue.push(mk(-(tot + lef[x][y] - 1), tot));

                while(-outQue.top().first < tot) {
                    add(sum, outQue.top().second, -1);
                    outQue.pop();
                }

                ll tmp = query(sum, max(1, tot - rig[x][y] + 1), tot);
                ans += tmp;
            }
        for(int i = 1; i <= tot; ++i) sum[i] = 0;
        while(!outQue.empty()) outQue.pop();
    }

    cout << ans << endl;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%s", graph[i] + 1);
    solve();
    return 0;
}