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 = 100000000;
int M;
int N;
int d[256][256] = { 0 };
char maze[256][256] = {0};
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };
int sx, sy;
int gx, gy;
int  BFS()
{
    queue<P> que;
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            d[i][j] = INF;
        }
    }
    que.push(P(sx, sy));
    d[sx][sy] = 0;
    while (que.size())
    {
        P p = que.front();
        que.pop();
        if (p.first == gx && p.second == gy)break;

        for (int i = 0; i < 4; i++)
        {
            int nx = p.first + dx[i], ny = p.second + dy[i];
            if ( maze[nx][ny] == 'G' || 0 <= nx && nx < N && 0 <= 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] + 1;
            }
        }
    }
    return d[gx][gy];

}
void solve( )
{
    int res = BFS();
    cout << endl << res << endl;
}
int main()
{
    ifstream filein("data.txt");
    int i = 0;
    while (!filein.eof())
    {
        filein.getline(maze[i], 100);
        cout << maze[i++] << endl;
    }
    int m = 0, n = 0;
    while (maze[0][n])n++;
    while (maze[m][0])
    {
        for (int i = 0; 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 0;
}

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[4]={0,1,0,-1};
        int dy[4]={1,0,-1,0};
        bool res=false;
        queue<dot> q;
        q.push(dot(sx,sy));
        while(q.size()!=0){
            dot d=q.front();
            q.pop();
            
            if(grid[d.x][d.y]==2)res=true;
            grid[d.x][d.y]=3;
            for(int i=0;i<4;i++){
                int nx=d.x+dx[i];
                int ny=d.y+dy[i];
                if(nx>=0 && ny>=0 && nx<m && ny<n && grid[nx][ny]!=3 && grid[nx][ny]!=0){
                    q.push(dot(nx,ny));
                }
             }
         }
        
        return res;
         
         
    }
    int numIslandCities(vector<vector<int>> &grid) {
        // Write your code here
        int island_count=0;
        int m=grid.size();
        int n=grid[1].size();
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++){
                if(grid[i][j]==3||grid[i][j]==0){
                    continue;
                }
                if(bfs(grid,i,j,m,n))
                {
                    island_count++;
                }
            }
        return island_count;
    }
 
};