题目传送门
1 /*
2 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。
3 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-
4 */
5 /************************************************
6 Author :Running_Time
7 Created Time :2015-8-4 9:02:06
8 File Name :POJ_3984.cpp
9 ************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 #define lson l, mid, rt << 1
31 #define rson mid + 1, r, rt << 1 | 1
32 typedef long long ll;
33 const int MAXN = 10;
34 const int INF = 0x3f3f3f3f;
35 const int MOD = 1e9 + 7;
36 int a[MAXN][MAXN];
37 bool vis[MAXN][MAXN];
38 int dir[MAXN][MAXN];
39 int step[MAXN][MAXN];
40 int dx[4] = {-1, 1, 0, 0};
41 int dy[4] = {0, 0, -1, 1};
42 int n = 4, m = 4;
43
44 bool judge(int x, int y) {
45 if (x < 0 || x > n || y < 0 || y > m || a[x][y] == 1) return false;
46 return true;
47 }
48
49 void print_path(void) {
50 int x = n, y = m; vector<pair<int, int> > ans;
51 while (dir[x][y] != -1) {
52 ans.push_back (make_pair (x, y));
53 int px = x, py = y;
54 x -= dx[dir[px][py]]; y -= dy[dir[px][py]];
55 }
56 int sz = (int) ans.size ();
57 printf ("(0, 0)
");
58 for (int i=sz-1; i>=0; --i) {
59 printf ("(%d, %d)
", ans[i].first, ans[i].second);
60 }
61 }
62
63 void BFS(void) {
64 memset (vis, false, sizeof (vis));
65 memset (step, INF, sizeof (step));
66 memset (dir, -1, sizeof (dir));
67 queue<pair<int, int> > Q; Q.push (make_pair (0, 0)); vis[0][0] = true;
68 step[0][0] = 0;
69 while (!Q.empty ()) {
70 int x = Q.front ().first, y = Q.front ().second; Q.pop ();
71 for (int i=0; i<4; ++i) {
72 int tx = x + dx[i], ty = y + dy[i];
73 if (!judge (tx, ty)) continue;
74 if (vis[tx][ty] && step[tx][ty] <= step[x][y] + 1) continue;
75 if (tx == n && ty == m) {
76 dir[tx][ty] = i; print_path (); return ;
77 }
78 dir[tx][ty] = i; step[tx][ty] = step[x][y] + 1;
79 Q.push (make_pair (tx, ty)); vis[tx][ty] = true;
80 }
81 }
82 }
83
84 int main(void) { //POJ 3984 迷宫问题
85 for (int i=0; i<5; ++i) {
86 for (int j=0; j<5; ++j) {
87 scanf ("%d", &a[i][j]);
88 }
89 }
90 BFS ();
91
92 return 0;
93 }