題目: 給定一個大小為N*M的迷宮. 迷宮由通道和墻壁組成, 每一步可以向鄰接的上下左右四格的通道移動.

請求出從起點到終點所需的最小步數. 請注意, 本題假定從起點一定可以移動到終點.

使用寬度優先搜索算法(DFS), 依次遍歷迷宮的四個方向, 當有可以走且未走過的方向時, 移動并且步數加一.

時間復雜度取決于迷宮的狀態數, O(4*M*N)=O(M*N).

代碼:

/*
 * main.cpp
 *
 *  Created on: 2014.7.17
 *      Author: spike
 */

/*eclipse cdt, gcc 4.8.1*/

#include <stdio.h>
#include <limits.h>

#include <utility>
#include <queue>

using namespace std;

class Program {
	static const int MAX_N=20, MAX_M=20;
	const int INF = INT_MAX>>2;
	typedef pair<int, int> P;

	char maze[MAX_N][MAX_M+1] = {
			"#S######.#",
			"......#..#",
			".#.##.##.#",
			".#........",
			"##.##.####",
			"....#....#",
			".#######.#",
			"....#.....",
			".####.###.",
			"....#...G#"
	};
	int N = 10, M = 10;
	int sx=0, sy=1; //起點坐標
	int gx=9, gy=8; //重點坐標

	int d[MAX_N][MAX_M];

	int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; //四個方向移動的坐標

	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 (0<=nx&&nx<N&&0<=ny&&ny<M&&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];
	}
public:
	void solve() {
		int res = bfs();
		printf("result = %d\n", res);
	}
};


int main(void)
{
	Program P;
	P.solve();
    return 0;
}