Zbazi in Zeydabad CodeForces - 628E 树状数组

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.

Example

Input

4 4

zzzz

zzz.

.z..

zzzz

Output

16

Input

1 4

z.z.

Output

2

Input

2 2

zz

zz

Output

5

在一个正方形内，第一行和最后一行和反对角线全是’z’字符，则算满足条件，求有几个这样的正方形。

预处理每个点左边，右边，左下可以延伸的长度，暴力的思路是枚举每个点作为右上角，正方形的最大边长是第一条边和对角线的最小值，然后在范围内判断对角线上的每个点向右延伸的长度是否满足条件，每有一个点满足就是符合的图形。

由于不同的正方形对角线可能共线，有重叠的部分，判断区间内个数的操作可以用线段树优化。

预处理时记录每条线段的右端点，然后横坐标从右往左读取线段，每个读入的线段左端点归属于唯一的一条对角线i+k，对角线i+k上i位置出发的线条计数+1，而从右往左读取保证了此时已读取的线条右端点一定在横坐标j右边。接着区间查询当前枚举的顶点所在的对角线上，左端点在对角线上且在区间[i,i+len-1]中，右端点>=j(已经读取的一定满足)的线段。

#include <iostream>
#include <stdio.h>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <math.h>
#include <iterator>
#include <string.h>
using namespace std;
typedef long long ll;
int mo[4][2]={0,1,1,0,0,-1,-1,0};
const int MAXN=0x3f3f3f3f;
const int sz=3005;
int n,m;
ll ans;
char mp[sz][sz];
ll cnt[2*sz][sz];//储存量有i+j
int l[sz][sz],r[sz][sz],dia[sz][sz],ed[sz][sz];

void add(int num,int j){
while(j<=n){
cnt[num][j]++;
j+=j&(-j);
}
}

ll sum(int num,int j){
int re=0;
while(j>0){
re+=cnt[num][j];
j-=j&(-j);
}
return re;
}

int main()
{
//freopen("r.txt","r",stdin);
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=1;i<=n;i++){
scanf("%s",mp[i]+1);
}
ans=0;
memset(l,0,sizeof(l));
memset(r,0,sizeof(r));
memset(cnt,0,sizeof(cnt));
memset(dia,0,sizeof(dia));
memset(ed,0,sizeof(ed));
for(int i=n;i>=1;i--){
for(int j=1;j<=m;j++){
if(mp[i][j]=='z'){
l[i][j]=l[i][j-1]+1;
}else{
l[i][j]=0;
}
}
for(int j=m;j>=1;j--){
if(mp[i][j]=='z'){
r[i][j]=r[i][j+1]+1;
if(mp[i][j+1]!='z'){
ed[i][j]=1;
}
}else{
r[i][j]=0;
}
}
for(int j=1;j<=m;j++){
if(mp[i][j]=='z'){
dia[i][j]=dia[i+1][j-1]+1;
}else{
dia[i][j]=0;
}
}
}
for(int j=m;j>=1;j--){
for(int i=1;i<=n;i++){
if(ed[i][j]){//ed表示有在j处结尾的线段
int k=j;
for(;mp[i][k]=='z';k--){
add(i+k,i);//i+k为对角线编号，线段起点，i是该线段在对角线上属于的行数，用于确定范围
//现在加入的线段和已经加入的线段右端点都大于等于J，保证满足要求
}
}
}
for(int i=1;i<=n;i++){
int len=min(l[i][j],dia[i][j]);
ll t=(sum(i+j,i+len-1)-sum(i+j,i-1));
ans+=t;
}
}
cout<<ans<<endl;
}
return 0;
}