E. Zbazi in Zeydabad

题目连接：

http://www.codeforces.com/contest/628/problem/D

Description

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 use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/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.

Sample Input

4 4
zzzz
zzz.
.z..
zzzz

Sample Output

16

Hint

题意

给你一个n*m的矩阵，然后问你里面有多少个z

题解：

数据结构

首先我们对于每一个点维护一个l[i][j],r[i][j]，表示这个点往左边最多延伸多少，往右边最多延伸多少

我们去维护每一个对角线就好了，我们从左下角跑到右上角

我们update(i,1)表示这个位置当前是可以的，他对于上面的i+r[i][j]-1个都是可以更新的

他这个点的贡献就是query(i)-query(i-len)，len = l[i][j]和i-last的最小值，显然他可以和下面len个进行组合

然后这样去跑就好了

我讲的比较抽象，还是看代码吧……

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e3+5;
int a[maxn],n,m;
char s[maxn][maxn];
int l[maxn][maxn],r[maxn][maxn];
vector<int>over[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    x++;
    for(int i=x;i<maxn;i+=lowbit(i))
        a[i]+=val;
}
int query(int x)
{
    x++;
    int ans=0;
    for(int i=x;i;i-=lowbit(i))
        ans+=a[i];
    return ans;
}
long long solve(int x,int y)
{
    memset(a,0,sizeof(a));
    int L = min(m-y,x+1);
    for(int i=0;i<maxn;i++)over[i].clear();
    int last=-1;
    long long ret = 0;
    for(int i=0;i<L;i++)
    {
        int nx = x-i;
        int ny = y+i;
        if(s[nx][ny]=='.')
            last=i;
        else{
            update(i,1);over[i+r[nx][ny]-1].push_back(i);
            int len = min(l[nx][ny],i-last);
            ret+=query(i)-query(i-len);
        }
        for(int j=0;j<over[i].size();j++)
            update(over[i][j],-1);
    }
    return ret;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
        scanf("%s",s[i]);
    for(int i=0;i<n;i++)
    {
        l[i][0]=s[i][0]=='z';
        for(int j=1;j<m;j++)
            if(s[i][j]=='z')l[i][j]=l[i][j-1]+1;
        r[i][m-1]=s[i][m-1]=='z';
        for(int j=m-2;j>=0;j--)
            if(s[i][j]=='z')r[i][j]=r[i][j+1]+1;
    }
    long long ans = 0;
    for(int i=0;i<n;i++)
        ans+=solve(i,0);
    for(int i=1;i<m;i++)
        ans+=solve(n-1,i);
    printf("%lld\n",ans);
}