[Blue Bridge Cup] [2018 Ninth Zhenti] global warming (BFS)

topic

Do you have a sea Zhang  N × N . "" N × N pixel photos, represents the ocean, "#" represents the land, as follows:

.......
.##....
.##....
....##.
..####.
...###. ....... 

The "up and down" on the four directions together a piece of land consisting of an island, for example, the figure had  2 two islands.

Due to the rising sea levels caused by global warming, scientists predict the next few decades, the islands edge a pixel range will be submerged.

Specifically, (four adjacent pixels in the vertical and horizontal marine) If a pixel and adjacent land ocean, it will be submerged.

For example in the figure above sea will become the future looks as follows:

.......
.......
.......
.......
....#..
.......
.......

Please work: According to the scientists predict, the number of photos in the islands will be completely submerged.

Input Format

The first row contains an integer N.

The following  N N rows  N N column containing a character from the "#" and "." Consisting of  N × N represents the character N × N matrix, representing a sea picture, "#" land "." Represents the ocean.

Photos ensure that the first  1 row 1,  1 1, the first  N N th,  N N columns of pixels are ocean.

Output Format

An integer that represents the answer.

data range

1≤N≤10001≤N≤1000

Sample Input 1:

7
.......
.##....
.##....
....##.
..####.
...###. ....... 

Output Sample 1:

1

Sample Input 2:

9
.........
.##.##...
.#####...
.##.##...
.........
.##.#....
.#.###... .#..#.... ......... 

Output Sample 2:

思路

• To know the island where, you first need to know the pixel map that communicates using BFS traversal islands
• Determine whether a pixel is a border island, to see whether there is "·" It's down about the four elements

Need to pay attention to the point

• While traversing the island need to determine: 1) whether the point has been out of bounds 2) whether the point has been visited 3) The point is the "#" or "*"
• Control direction, creating dx [] = {-1,0,1,0}, dy [] = {0, 1, 0, -1} two arrays, four times with a for loop, the upper left to lower right turn order.

Code

 

#include<iostream>
#include<queue>
#define MAXN 1001

using namespace std;
int n;
char img[MAXN][MAXN];
bool vis[MAXN][MAXN];
int dx[] = { -1,0,1,0 }, dy[] = { 0, 1, 0, -1 };
/*
7
.......
.##....
.##....
....##.
..####.
...###.
.......
*/

void bfs(int i, int j, int &sum, int &bound) {
    vis[i][j] = true;
    queue<pair<int, int>>q;
    pair<int, int> p(i, j);
    q.push(p);
    while (!q.empty()) {
        auto l = q.front();
        q.pop();
        sum++;
        bool is_bound = false;
        for (int k = 0; k < 4; k++) {
            int x = l.first + dx[k], y = l.second + dy[k];
            if (x > n || y > n || x < 0 || y < 0 || vis[x][y]==true)continue;
            else if (img[x][y] == '#') {
                vis[x][y] = true;
                p = make_pair(x, y);
                q.push(p);
            }
            else if (img[x][y] == '.') {
                is_bound = true;
            }
            
        }
        if (is_bound == true)bound++;
    }
}


int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> img[i][j];
        }
    }
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (img[i][j] == '#' && vis[i][j] == false) {
                int sum = 0, bound = 0;
                bfs(i, j, sum, bound);
                if (sum == bound)count++;
            }
        }
    }
    cout << count;
    return 0;
}