链接：

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 
    
     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\n");    else        printf("NO\n");    return 0;}