链接：
https://codeforces.com/contest/1182/problem/B
题意：
You have a given picture with size w×h. Determine if the given picture has a single "+" shape or not. A "+" shape is described below:
A "+" shape has one center nonempty cell.
There should be some (at least one) consecutive non-empty cells in each direction (left, right, up, down) from the center. In other words, there should be a ray in each direction.
All other cells are empty.
Find out if the given picture has single "+" shape.
思路：
遍历每一个点，当某个点的上下左右都为时。向四个方向扩展，记录的个数。之后结束遍历。
当某个点可以形成+且的个数等于所有*的个数的时候，YES。
代码：
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 1e3 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;
char pi[MAXN][MAXN];
int main()
{
scanf("%d %d", &n, &m);
getchar();
int sum = 0;
for (int i = 1;i <= n;i++)
{
for (int j = 1;j <= m;j++)
{
scanf("%c", &pi[i][j]);
if (pi[i][j] == '*')
sum++;
}
getchar();
}
bool flag = false;
for (int i = 2;i <= n-1;i++)
{
for (int j = 2;j <= m-1;j++)
{
if (pi[i][j]=='*'&&pi[i-1][j]=='*'&&pi[i][j+1]=='*'&&pi[i+1][j]=='*'&&pi[i][j-1]=='*')
{
flag = true;
sum--;
int tx, ty;
tx = i-1, ty = j;
while (tx >= 1 && pi[tx][ty] == '*')
tx--, sum--;
tx = i, ty = j+1;
while (ty <= m && pi[tx][ty] == '*')
ty++, sum--;
tx = i+1, ty = j;
while (tx <= n && pi[tx][ty] == '*')
tx++, sum--;
tx = i, ty = j-1;
while (ty >= 1 && pi[tx][ty] == '*')
ty--, sum--;
}
if (flag)
break;
}
cerr << endl;
if (flag)
break;
}
if (flag && sum == 0)
printf("YES
");
else
printf("NO
");
return 0;
}