思路：先预处理一下每个元素之后会有多少个连续的’*’,时间O(n^3），然后对于每一个元素暴力模拟一下，时间O
(n^3)，细节见代码。
Code:
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<bitset>
#define pii pair<int,int>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;
int lst[505][505];
int Mod = 1e9 + 7;
int n, m;ll sum = 0;
int dp[505][505];
void check(int x,int y)
{
int g = 0;int f = 0;
while (1)
{
if (x + g > n || y + g > m||y-g<1)break;
if ( dp[x+g][y-g]<1+2*g)f = 1;
g++;
if (f)break;
sum++;
}
}
int main()
{
FAST;
int t;cin >> t;
while (t--)
{
cin >> n >> m;
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
char g;cin >> g;
if (g == '*')lst[i][j] = 1;
else lst[i][j] = 0;
}
}
sum = 0;
memset(dp, 0, sizeof(dp));
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
int su = 0;
for (int k = j;k <= m;k++)
{
if (lst[i][k] == 1)su++;
else break;
}
dp[i][j] = su;
}
}
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
if (lst[i][j])check(i, j);
}
}
cout << sum << endl;
}
}