BFS小记
题目：求从S走到G点所需步数
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#S######.# |
此代码BFS遍历顺序：
宽度优先搜索关键点：
- 所用容器：队列
- 更新：取队列的队首
- BFS终止条件：队列不为空
- 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：
1 #include <iostream> 2 #include <fstream> 3 #include <queue> 4 #include <iomanip> 5 6 using namespace std; 7 typedef pair<int, int> P; 8 const int INF = 100000000; 9 int M; 10 int N; 11 int d[256][256] = { 0 }; 12 char maze[256][256] = {0}; 13 int dx[4] = { 1, 0, -1, 0 }; 14 int dy[4] = { 0, 1, 0, -1 }; 15 int sx, sy; 16 int gx, gy; 17 int BFS() 18 { 19 queue<P> que; 20 for (int i = 0; i < N; i++) 21 { 22 for (int j = 0; j < M; j++) 23 { 24 d[i][j] = INF; 25 } 26 } 27 que.push(P(sx, sy)); 28 d[sx][sy] = 0; 29 while (que.size()) 30 { 31 P p = que.front(); 32 que.pop(); 33 if (p.first == gx && p.second == gy)break; 34 35 for (int i = 0; i < 4; i++) 36 { 37 int nx = p.first + dx[i], ny = p.second + dy[i]; 38 if ( maze[nx][ny] == 'G' || 0 <= nx && nx < N && 0 <= ny && ny < M && maze[nx][ny] != '#' && maze[nx][ny] == '.' && d[nx][ny] == INF ) 39 { 40 que.push(P(nx, ny)); 41 d[nx][ny] = d[p.first][p.second] + 1; 42 } 43 } 44 } 45 return d[gx][gy]; 46 47 } 48 void solve( ) 49 { 50 int res = BFS(); 51 cout << endl << res << endl; 52 } 53 int main() 54 { 55 ifstream filein("data.txt"); 56 int i = 0; 57 while (!filein.eof()) 58 { 59 filein.getline(maze[i], 100); 60 cout << maze[i++] << endl; 61 } 62 int m = 0, n = 0; 63 while (maze[0][n])n++; 64 while (maze[m][0]) 65 { 66 for (int i = 0; i < n; i++) 67 { 68 if (maze[m][i] == 'S') 69 { 70 sx = m; 71 sy = i; 72 } 73 else if (maze[m][i] == 'G') 74 { 75 gx = m; 76 gy = i; 77 } 78 } 79 m++; 80 } 81 cout << "S:" << sx << " " << sy << " " << endl; 82 cout << "G:" << gx << " " << gy << " " << endl; 83 cout << "m:" << m << endl; 84 cout << "n:" << n << endl; 85 M = m; 86 N = n; 87 solve(); 88 return 0; 89 }
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/
1 class Solution { 2 public: 3 /** 4 * @param grid: an integer matrix 5 * @return: an integer 6 */ 7 class dot{ 8 public: 9 int x; 10 int y; 11 dot(int xx,int yy){ 12 this->x=xx; 13 this->y=yy; 14 15 } 16 17 }; 18 19 bool bfs(vector<vector<int>> &grid,int sx,int sy,int m,int n){ 20 int dx[4]={0,1,0,-1}; 21 int dy[4]={1,0,-1,0}; 22 bool res=false; 23 queue<dot> q; 24 q.push(dot(sx,sy)); 25 while(q.size()!=0){ 26 dot d=q.front(); 27 q.pop(); 28 29 if(grid[d.x][d.y]==2)res=true; 30 grid[d.x][d.y]=3; 31 for(int i=0;i<4;i++){ 32 int nx=d.x+dx[i]; 33 int ny=d.y+dy[i]; 34 if(nx>=0 && ny>=0 && nx<m && ny<n && grid[nx][ny]!=3 && grid[nx][ny]!=0){ 35 q.push(dot(nx,ny)); 36 } 37 } 38 } 39 40 return res; 41 42 43 } 44 int numIslandCities(vector<vector<int>> &grid) { 45 // Write your code here 46 int island_count=0; 47 int m=grid.size(); 48 int n=grid[1].size(); 49 50 for(int i=0;i<m;i++) 51 for(int j=0;j<n;j++){ 52 if(grid[i][j]==3||grid[i][j]==0){ 53 continue; 54 } 55 if(bfs(grid,i,j,m,n)) 56 { 57 island_count++; 58 } 59 } 60 return island_count; 61 } 62 63 };
