Description     定义一个二维数组：

int maze[5][5] = {
0,1,0,0,0,
0,1,0,1,0,
0,0,0,0,0,
0,1,1,1,0,
0,0,0,1,0,
};
它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。

Input     一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。

Output     左上角到右下角的最短路径，格式如样例所示。

Sample Input

0 1 0 0 0
0 1 0 1 0
0 0 0 0 0
0 1 1 1 0
0 0 0 1 0

Sample Output

(0, 0)
(1, 0)
(2, 0)
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(3, 4)
(4, 4)

思路：定义一个数组dis用来表示（0，0）点到当前点要做几步，用BFS走一遍，然后从（4,4）点往回走，先把（4,4）入栈，令dis<4,4>为a【从（0,0）到（4,4）要走的步数】，令当前点为（4，4），①则走到则当前点的上一点需要的步数为a-1，②遍历当前点的周围四方向【不越界】，③找出该点入栈，④把该点作为当前点，重复上述①②③④，直到回到（0,0）；

本题用DFS和BFS均可

已Accept代码【c++提交】

BFS☟

0MS 144K

#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;

int map[6][6];
int dis[6][6];
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
typedef struct node{
	int x, y;
	node() {}
	node(int x, int y) : x(x), y(y) {}
}node;

void BFS() {
	queue <node> Q;
	memset(dis, 100, sizeof(dis));
	Q.push(node(0, 0));
	dis[0][0] = 0;
	while(!Q.empty()) {
		node k = Q.front();
		Q.pop();
		int x = k.x;
		int y = k.y;
		for(int i = 0; i < 4; i++) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if(nx < 0 || nx >= 5 || ny < 0 || ny >= 5 || map[nx][ny] == 1)
				continue;
			if(dis[nx][ny] > dis[x][y] + 1) {
				dis[nx][ny] = dis[x][y] + 1;
				Q.push(node(nx, ny));
			}
		}
	}
	stack <node> S;
	int x = 4, y = 4;
	S.push(node(x, y));
	while(1) {
		if(x == 0 && y == 0)
			break;
		for(int i = 0; i < 4; i++) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if(nx < 0 || nx >= 5 || ny < 0 || ny >= 5)
				continue;
			if(dis[x][y] == dis[nx][ny] + 1) {
				x = nx;
				y = ny;
				S.push(node(x, y));
			}
		}
	}
	while(!S.empty()) {
		node a = S.top();
		S.pop();
		printf("(%d, %d)\n", a.x, a.y);
	}
}

int main() {
	for(int i = 0; i < 5; i++)
		for(int j = 0; j < 5; j++) 
			scanf("%d", &map[i][j]);
	BFS();
	return 0;
}

DFS☟

0MS 164K

#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int map[6][6];
int dis[7][7];
int n[18];
int m[18];
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};

void dfs(int sc, int sr, int total) {
	if(sc < 1 || sc > 5 || sr < 1 || sr > 5)
		return ;
	if(map[sc][sr] == 1)
		return ;
	if(total  + 1 >= dis[sc][sr])
		return ;
	dis[sc][sr] = total  + 1 ;
	map[sc][sr] = 1;
	for(int i = 0; i < 4; i++) {
		int nx = sc + dx[i];
		int ny = sr + dy[i];
		dfs(nx, ny, dis[sc][sr]);
	}
	map[sc][sr] = 0;
	return ;
}

void Scanf_a() {
	memset(dis, 0x3f, sizeof(dis));
	for(int i = 1; i < 6; i++)
		for(int j = 1; j < 6; j++)
			scanf("%d", &map[i][j]);
}

void Printf_a() {
	int x = 5, y = 5, q = 0;
	int current = dis[x][y];
	n[q] = x;
	m[q++] = y;
	while(x != 1 || y !=1) {
		for(int i = 0; i < 4; i++) {
			if(dis[x + dx[i]][y + dy[i]] == current - 1) {
				x = x + dx[i];
				y = y + dy[i];
				n[q] = x;
				m[q++] = y;
				current--;
			}
		}
	}
	for(int j = dis[5][5] - 1; j >= 0; j--)
		printf("(%d, %d)\n", n[j] - 1, m[j] - 1);
}

int main(){
	Scanf_a();
	dfs(1, 1, 0);
	Printf_a();
	system("pause");
	return 0;
}