标签：strong ati atom upper min you ica data- ranch
You have a given picture with size w×hw×h. Determine if the given picture has a single "+" shape or not. A "+" shape is described below:
Find out if the given picture has single "+" shape.
Input
The first line contains two integers hh and ww (1≤h1≤h, w≤500w≤500) — the height and width of the picture.
The ii-th of the next hh lines contains string sisi of length ww consisting "." and "*" where "." denotes the empty space and "*" denotes the non-empty space.
Output
If the given picture satisfies all conditions, print "YES". Otherwise, print "NO".
You can output each letter in any case (upper or lower).
Examples
5 6
......
..*...
.****.
..*...
..*...
YES
3 5
..*..
****.
.*...
NO
7 7
.......
...*...
..****.
...*...
...*...
.......
.*.....
NO
5 6
..**..
..**..
******
..**..
..**..
NO
3 7
.*...*.
***.***
.*...*.
NO
5 10
..........
..*.......
.*.******.
..*.......
..........
NO
Note
In the first example, the given picture contains one "+".
In the second example, two vertical branches are located in a different column.
In the third example, there is a dot outside of the shape.
In the fourth example, the width of the two vertical branches is 22.
In the fifth example, there are two shapes.
In the sixth example, there is an empty space inside of the shape.
#define TLE std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); cout.precision(10);
char ch[500+5][500+5];
int main()
{
TLE;
int n,m,x,y,flag=1;
while(cin>>n>>m)
{
for(int i=0; i<n; i++)
for(int j=0; j<m; j++)
{
cin>>ch[i][j];
}
int ans=0,xx=0;
//debug(ch,n,m);
int nx , ny;
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(ch[i][j]==‘*‘ && ch[i][j-1]==‘*‘ && ch[i][j+1]==‘*‘ && ch[i-1][j]==‘*‘ && ch[i+1][j]==‘*‘ && j-1>=0 && j+1<m && i-1>=0 && i+1<n)
{
ch[i][j] = ‘.‘;
for(int kkkk=0; kkkk<4; kkkk++)
{
nx = dir[kkkk][0] + i ;
ny = dir[kkkk][1] + j ;
while(ch[nx][ny]==‘*‘)
{
ch[nx][ny] = ‘.‘;
nx += dir[kkkk][0];
ny += dir[kkkk][1];
}
}
xx=1;break;
}
}
if(xx) break;
}
int kk=0;
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
if(ch[i][j]==‘*‘)
{
kk=1;
break;
}
}
if(kk) break;
}
if(xx==0 || kk )
cout<<"NO"<<endl;
else
cout<<"YES"<<endl;
}
ok;
}
G - Plus from Picture CodeForces - 1182B
标签：strong ati atom upper min you ica data- ranch
原文地址：https://www.cnblogs.com/Shallow-dream/p/11618769.html
