DFS-C++实现水洼数统计

DFS-C++实现水洼数统计

#include <iostream>
using namespace std;
const int num = 100;
string a[1000][1000];//草地矩阵
int visited[num][num];//访问数组
int n , m;//n行，m列
int dx[8] = {-1 , -1 , 0 , 1 , 1 , 1 , 0 , -1};//8个方向的x坐标变化
int dy[8] = {0 , 1 , 1 , 1 , 0 , -1 , -1 , -1};//8个方向的y坐标变化
int Count = 0;
void init()
{
    cout<<"请输入水洼的行列数:"<<endl;
    cin>>n>>m;
    cout<<"请输入草地情况:"<<endl;
    for(int i = 1 ; i <= n ; i++)
    {
        for(int j = 1 ; j <= m; j++)
        {
            cin>>a[i][j];
            visited[i][j] = 0;
        }
    }
}

void dfs(int x , int y)
{
    visited[x][y] = 1;
    for(int i = 0 ; i < 8 ; i++)
    {
        int newX = x + dx[i];
        int newY = y + dy[i];

        if(newX >= 1 && newX <= n && newY >= 1 && newY <= m && visited[newX][newY] == 0 && a[newX][newY] == "X")
        {//如果新的X，Y坐标为越界并且未被访问过且为W就去递归
            visited[newX][newY] = 1;
            dfs(newX , newY);
        }
    }

}
int main()
{
    init();
    for(int i = 1 ; i <= n ; i++)
    {
        for(int j = 1 ; j <= m ; j++)
        {
            if(a[i][j] == "X" && visited[i][j] == 0)
            {
                dfs(i , j);
                Count++;
            }
        }
    }
    cout<<"水洼的个数为:"<<Count<<endl;
    return 0;
}