迷宫的最短路径 代码(C++)

迷宫的最短路径 代码(C++)

本文地址: http://blog.csdn.net/caroline_wendy
题目: 给定一个大小为N*M的迷宫. 迷宫由通道和墙壁组成, 每一步可以向邻接的上下左右四格的通道移动.

请求出从起点到终点所需的最小步数. 请注意, 本题假定从起点一定可以移动到终点.

使用宽度优先搜索算法(DFS), 依次遍历迷宫的四个方向, 当有可以走且未走过的方向时, 移动并且步数加一.

时间复杂度取决于迷宫的状态数, O(4*M*N)=O(M*N).

代码:

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/*

* 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; //起点坐标

int gx=9, gy=8; //重点坐标

int d[MAX_N][MAX_M];

int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; //四个方向移动的坐标

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;

}

输出:

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result = 22