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Square Detector20 points

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You want to write an image detection system that is able to recognize different geometric shapes. In the first version of the system you settled with just being able to detect filled squares on a grid.

You are given a grid of N×N square cells. Each cell is either white or black. Your task is to detect whether all the black cells form a square shape.

Input

The first line of the input consists of a single number T, the number of test cases.

Each test case starts with a line containing a single integer N. Each of the subsequent N lines contain N characters. Each character is either "." symbolizing a white cell, or "#" symbolizing a black cell. Every test case contains at least one black cell.

Output

For each test case printYESif all the black cells form a completely filled square with edges parallel to the grid of cells. Otherwise printNO.

Constraints

1 ≤ T ≤ 20 1 ≤ N ≤ 20

Example

Test cases 1 and 5 represent valid squares. Case 2 has an extra cell that is outside of the square. Case 3 shows a square not filled inside. And case 4 is a rectangle but not a square.

Example input · Download

Example output · Download

5
4
..##
..##
....
....
4
..##
..##
#...
....
4
####
#..#
#..#
####
5
#####
#####
#####
#####
.....
5
#####
#####
#####
#####
#####

Case #1: YES
Case #2: NO
Case #3: NO
Case #4: NO
Case #5: YES

#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <queue>

using namespace std;
#define INF 0x7fffffff
#define maxm 1001
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i = 0; i < (n); i++)
#define re return
#define fi first
#define se second
#define sz(x) ((int) (x).size())
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define sqrt(x) sqrt(abs(x))
#define y0 y3487465
//#define y1 y8687969
#define fill(x,y) memset(x,y,sizeof(x))

typedef vector<int> vi;
typedef long long ll;
typedef long double ld;
typedef double D;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef vector<vi> vvi;

template<class T> T abs(T x) { re x > 0 ? x : -x; }

const int maxn = 10005;

int n, m, t, s, k, x;
//string s;
int a[maxn][maxn];
char str[maxn];
int main(){
    int cas = 1;
    scanf("%d", &t);
    while (t--){
        memset(a, 0, sizeof a);
        scanf("%d", &n);
        for (int i = 0; i < n; i++){
            scanf("%s", str);
            for (int j = 0; j < n; j++)
            if (str[j] == '#')a[i][j] = 1;
        }
        int x1 = 21, y1 = 21, x2 = -1, y2 = -1;
        for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
        if (a[i][j])x1 = i, y1 = j, i = n, j = n;
        for (int i = n - 1; i >= 0; i--)
        for (int j = n - 1; j >= 0; j--)
        if (a[i][j])x2 = i, y2 = j, i = -1, j = -1;
        s = 0, k = 0;
        for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++){
            if (i >= x1&&j >= y1&&i <= x2&&j <= y2&&a[i][j])k++;
            if (a[i][j])s++;
        }
        printf("Case #%d: ", cas++);
        if (s == k&&s == (x2 - x1 + 1)*(y2 - y1 + 1)&&(x2-x1+1)==(y2-y1+1))printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}