BFS小记

题目：求从S走到G点所需步数

#S######.#
......#..#
.#.##.##.#
.#........
##.##.####
....#....#
.#######.#
....#.....
.####.###.
....#...G#

此代码BFS遍历顺序：

宽度优先搜索关键点：

1. 所用容器：队列
2. 更新：取队列的队首
3. BFS终止条件：队列不为空
4. 入队条件：判断边界和障碍。  注：条件满足后可追加处理代码

下面贴上代码：

 #include <iostream>
 #include <fstream>
 #include <queue>
 #include <iomanip>

 using namespace std;
 typedef pair<int, int> P;
 const int INF = ;
 int M;
 int N;
 int d[][] = {  };
 char maze[][] = {};
 int dx[] = { , , -,  };
 int dy[] = { , , , - };
 int sx, sy;
 int gx, gy;
 int  BFS()
 {
     queue<P> que;
     for (int i = ; i < N; i++)
     {
         for (int j = ; j < M; j++)
         {
             d[i][j] = INF;
         }
     }
     que.push(P(sx, sy));
     d[sx][sy] = ;
     while (que.size())
     {
         P p = que.front();
         que.pop();
         if (p.first == gx && p.second == gy)break;

         for (int i = ; i < ; i++)
         {
             int nx = p.first + dx[i], ny = p.second + dy[i];
             if ( maze[nx][ny] == 'G' ||  <= nx && nx < N &&  <= ny && ny < M && maze[nx][ny] != '#' && maze[nx][ny] == '.' && d[nx][ny] == INF )
             {
                 que.push(P(nx, ny));
                 d[nx][ny] = d[p.first][p.second] + ;
             }
         }
     }
     return d[gx][gy];

 }
 void solve( )
 {
     int res = BFS();
     cout << endl << res << endl;
 }
 int main()
 {
     ifstream filein("data.txt");
     int i = ;
     while (!filein.eof())
     {
         filein.getline(maze[i], );
         cout << maze[i++] << endl;
     }
     int m = , n = ;
     while (maze[][n])n++;
     while (maze[m][])
     {
         for (int i = ; i < n; i++)
         {
             if (maze[m][i] == 'S')
             {
                 sx = m;
                 sy = i;
             }
             else if (maze[m][i] == 'G')
             {
                 gx = m;
                 gy = i;
             }
         }
         m++;
     }
     cout << "S:" << sx << "  " << sy << "  " << endl;
     cout << "G:" << gx << "  " << gy << "  " << endl;
     cout << "m:" << m << endl;
     cout << "n:" << n << endl;
     M = m;
     N = n;
     solve();
     return ;
 }

运行结果：

这是典型的BFS迷宫问题，也是最基本的BFS，需要熟练掌握，但并不困难。

2019/2/13更新

lintcode 897

传送门：https://www.lintcode.com/problem/island-city/

 class Solution {
 public:
     /**
      * @param grid: an integer matrix
      * @return: an integer
      */
     class dot{
     public:
        int x;
        int y;
        dot(int xx,int yy){
            this->x=xx;
            this->y=yy;

        }

      };

     bool bfs(vector<vector<int>> &grid,int sx,int sy,int m,int n){
         int dx[]={,,,-};
         int dy[]={,,-,};
         bool res=false;
         queue<dot> q;
         q.push(dot(sx,sy));
         while(q.size()!=){
             dot d=q.front();
             q.pop();

             if(grid[d.x][d.y]==)res=true;
             grid[d.x][d.y]=;
             for(int i=;i<;i++){
                 int nx=d.x+dx[i];
                 int ny=d.y+dy[i];
                 if(nx>= && ny>= && nx<m && ny<n && grid[nx][ny]!= && grid[nx][ny]!=){
                     q.push(dot(nx,ny));
                 }
              }
          }

         return res;

     }
     int numIslandCities(vector<vector<int>> &grid) {
         // Write your code here
         int island_count=;
         int m=grid.size();
         int n=grid[].size();

         for(int i=;i<m;i++)
             for(int j=;j<n;j++){
                 if(grid[i][j]==||grid[i][j]==){
                     continue;
                 }
                 if(bfs(grid,i,j,m,n))
                 {
                     island_count++;
                 }
             }
         return island_count;
     }

 };