| Time Limit: 1000MS | Memory Limit: 262144KB | 64bit IO Format: %I64d & %I64u |
Description
Fox Ciel has a board with n rows and n columns. So, the board consists of n × n cells. Each cell contains either a symbol ‘.‘, or a symbol ‘#‘.
A cross on the board is a connected set of exactly five cells of the board that looks like a cross. The picture below shows how it looks.
Ciel wants to draw several (may be zero) crosses on the board. Each cross must cover exactly five cells with symbols ‘#‘, and any cell with symbol ‘#‘ must belong to some cross. No two crosses can share a cell.
Please, tell Ciel if she can draw the crosses in the described way.
Input
The first line contains an integer n (3 ≤ n ≤ 100) — the size of the board.
Each of the next n lines describes one row of the board. The i-th line describes the i-th row of the board and consists of n characters. Each character is either a symbol ‘.‘, or a symbol ‘#‘.
Output
Output a single line with "YES" if Ciel can draw the crosses in the described way. Otherwise output a single line with "NO".
Sample Input
5
.#...
####.
.####
...#.
.....
YES
4
####
####
####
####
NO
6
.#....
####..
.####.
.#.##.
######
.#..#.
YES
6
.#..#.
######
.####.
.####.
######
.#..#.
NO
3
...
...
...
YES
Hint
In example 1, you can draw two crosses. The picture below shows what they look like.
In example 2, the board contains 16 cells with ‘#‘, but each cross contains 5. Since 16 is not a multiple of 5, so it‘s impossible to cover all.
Source
#include <iostream> #include <stdlib.h> #include <stdio.h> #include <algorithm> #include <string.h> #include <math.h> using namespace std; typedef long long ll; int main() { int i,j,n,sum,flag; char c[105][105]; while (cin>>n) { sum=0; for (i=0;i<n;i++) for (j=0;j<n;j++) { cin>>c[i][j]; if (c[i][j]==‘#‘) sum++; } for (i=0;i<n;i++) for (j=0;j<n;j++) { if (c[i][j]==‘#‘&&c[i+1][j]==‘#‘&&c[i+2][j]==‘#‘&&c[i+1][j+1]==‘#‘&&c[i+1][j-1]==‘#‘) { c[i][j]=‘.‘; c[i+1][j]=‘.‘; c[i+2][j]=‘.‘; c[i+1][j+1]=‘.‘; c[i+1][j-1]=‘.‘; sum-=5; } } //cout<<sum<<endl; if (sum==0) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }
原文：http://www.cnblogs.com/Ritchie/p/5424990.html