BFS(最短路+路径打印) POJ 3984 迷宫问题

题目传送门

 /*
     BFS：额，这题的数据范围太小了。但是重点是最短路的求法和输出路径的写法。
         dir数组记录是当前点的上一个点是从哪个方向过来的，搜索+，那么回溯-
 */
 /************************************************
 Author        :Running_Time
 Created Time  :2015-8-4 9:02:06
 File Name     :POJ_3984.cpp
 ************************************************/

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <sstream>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <list>
 #include <map>
 #include <set>
 #include <bitset>
 #include <cstdlib>
 #include <ctime>
 using namespace std;

 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 typedef long long ll;
 const int MAXN = ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int a[MAXN][MAXN];
 bool vis[MAXN][MAXN];
 int dir[MAXN][MAXN];
 int step[MAXN][MAXN];
 int dx[] = {-, , , };
 int dy[] = {, , -, };
 int n = , m = ;

 bool judge(int x, int y)    {
     if (x <  || x > n || y <  || y > m || a[x][y] == )   return false;
     return true;
 }

 void print_path(void)   {
     int x = n, y = m;   vector<pair<int, int> > ans;
     while (dir[x][y] != -) {
         ans.push_back (make_pair (x, y));
         int px = x, py = y;
         x -= dx[dir[px][py]];    y -= dy[dir[px][py]];
     }
     int sz = (int) ans.size ();
     printf ("(0, 0)\n");
     for (int i=sz-; i>=; --i) {
         printf ("(%d, %d)\n", ans[i].first, ans[i].second);
     }
 }

 void BFS(void)  {
     memset (vis, false, sizeof (vis));
     memset (step, INF, sizeof (step));
     memset (dir, -, sizeof (dir));
     queue<pair<int, int> > Q; Q.push (make_pair (, ));    vis[][] = true;
     step[][] = ;
     while (!Q.empty ()) {
         int x = Q.front ().first, y = Q.front ().second;    Q.pop ();
         for (int i=; i<; ++i) {
             int tx = x + dx[i], ty = y + dy[i];
             if (!judge (tx, ty))    continue;
             if (vis[tx][ty] && step[tx][ty] <= step[x][y] + )  continue;
             if (tx == n && ty == m) {
                 dir[tx][ty] = i;    print_path ();  return ;
             }
             dir[tx][ty] = i;    step[tx][ty] = step[x][y] + ;
             Q.push (make_pair (tx, ty));    vis[tx][ty] = true;
         }
     }
 }

 int main(void)    {     //POJ 3984 迷宫问题
     for (int i=; i<; ++i)    {
         for (int j=; j<; ++j)    {
             scanf ("%d", &a[i][j]);
         }
     }
     BFS ();

     return ;
 }