迷宫问题 poj3984

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 00 1 0 1 00 0 0 0 00 1 1 1 00 0 0 1 0

Sample Output

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

//BFS广搜然后在找到路径后根据Pre往前递归到-1， 然后输出， 熟悉队列原理即可

#include <iostream>#include <cmath>#include <string.h>#include <algorithm>#include <iomanip>#include <set>#include <vector>#include <map>#include <list>#include <stack>#include <queue>#include <deque>using namespace std;#define PI 3.1415926#define INF 0x7fffffff#define maxn 5struct node{int x, y;int Pre;}q[1000];int cnt;int Dx[4] = {0, 1, 0, -1};int Dy[4] = {1, 0, -1, 0};int T, n, m, step;int MAP[maxn][maxn];bool book[maxn][maxn];void Init();void BFS(int x, int y);void Print(int front){if(q[front].Pre == -1)return ;Print(q[front].Pre);cout << "(" << q[front].x << ", " << q[front].y <<")" << endl;}int main(){std::ios::sync_with_stdio(false);n = 5;Init();memset(book, 0, sizeof(book));cout << "(0, 0)" << endl;BFS(0, 0);cout << "(4, 4)" << endl;return 0;}void Init(){for(int i = 0; i < 5; i++){for(int j = 0; j < 5; j++){cin >> MAP[i][j];}}}void BFS(int x, int y){book[0][0] = 1;q[0].x = x;q[0].y = y;q[0].Pre = -1;int front = 0, rear = 1;while(front < rear){for(int i = 0; i < 4; i++){int dx = q[front].x + Dx[i];int dy = q[front].y + Dy[i];if(dx < 0 || dx > 4 || dy < 0 || dy > 4)continue;else if(book[dx][dy] == 0 && MAP[dx][dy] == 0){book[dx][dy] = 1;q[rear].x = dx;q[rear].y = dy;q[rear].Pre = front;rear++;cnt = rear;}if(dx == 4 && dy == 4){Print(front);return;}}front++;}return;}