POJ 3984 迷宫问题 bfs

题目链接：迷宫问题

天啦撸。最近怎么了。小bug缠身，大bug 不断。然这是我大腿第一次给我dbug。虽然最后的结果是。我............bfs入队列的是now..............

然后。保存路径的一种用的string 。一种用的数组。大同小异。根据就是我bfs 先搜到的绝壁就是步数最少的、

附代码：

pre 数组

 1 /*
 2  很简单的广搜。如果不是+路径输出的话。。
 3  保存路径。
 4  */
 5
 6 #include <stdio.h>
 7 #include <string.h>
 8 #include <iostream>
 9 #include <queue>
10 #include <string>
11 using namespace std;
12
13 struct Node {
14     int x, y;
15 }now, temp, ans;
16
17 int mp[10][10];
18 int vis[10][10];
19 Node que[210];
20 Node pre[210];
21
22 int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
23
24 bool check(Node a) {
25     if (a.x >= 0 && a.x < 5 && a.y >= 0 && a.y < 5 && !vis[a.x][a.y] && mp[a.x][a.y] == 0)
26         return true;
27     return false;
28 }
29
30 int head = 0, tail = 0;
31
32 void bfs() {
33     while(head < tail) {
34         now = que[head++];
35         if (now.x == 4 && now.y == 4) {
36             return;
37         }
38         for (int i=0; i<4; ++i) {
39             temp.x = now.x + dir[i][0];
40             temp.y = now.y + dir[i][1];
41
42             if (check(temp)) {
43                 que[tail++] = temp;
44                 vis[temp.x][temp.y] = 1;
45                 int id = temp.x * 5 + temp.y;
46                 pre[id].x = now.x, pre[id].y = now.y;
47             }
48         }
49     }
50 }
51
52 void outPre(int num) {
53     if (pre[num].x == -1 && pre[num].y == -1)
54         return;
55     else outPre(5*pre[num].x+pre[num].y);
56     cout << "(" << pre[num].x << ", " << pre[num].y << ")\n";
57 }
58
59 int main() {
60     head = 0, tail = 0;
61     memset(vis, 0, sizeof(vis));
62
63     for (int i=0; i<5; ++i) {
64         for (int j=0; j<5; ++j) {
65             cin >> mp[i][j];
66         }
67     }
68     now.x = 0, now.y = 0;
69     vis[0][0] = 1;
70     que[tail++] = now;
71     pre[0].x = -1, pre[0].y = -1;
72     bfs();
73
74     outPre(24);
75     cout << "(4, 4)\n";
76     return 0;
77 }

string

 1 /*
 2  很简单的广搜。如果不是+路径输出的话。想了一下 好像是dfs？好像不是啊。
 3
 4  */
 5
 6 #include <stdio.h>
 7 #include <string.h>
 8 #include <iostream>
 9 #include <queue>
10 #include <string>
11 using namespace std;
12
13 struct Node {
14     int x, y;
15     string ansStep;
16 }now, temp, ans;
17
18 int mp[10][10];
19 int vis[10][10];
20 Node que[210];
21
22 int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
23 bool check(Node a) {
24     if (a.x >= 0 && a.x < 5 && a.y >= 0 && a.y < 5 && !vis[a.x][a.y] && mp[a.x][a.y] == 0)
25         return true;
26     return false;
27 }
28
29 int head = 0, tail = 0;
30
31 void bfs() {
32     while(head < tail) {
33         now = que[head++];
34         if (now.x == 4 && now.y == 4) {
35             now.ansStep += ‘\0‘;
36             ans.ansStep = now.ansStep;
37             return;
38         }
39         for (int i=0; i<4; ++i) {
40             temp.x = now.x + dir[i][0];
41             temp.y = now.y + dir[i][1];
42
43             if (check(temp)) {
44                 vis[temp.x][temp.y] = 1;
45                 temp.ansStep = now.ansStep;
46                 temp.ansStep += temp.x + ‘0‘;
47                 temp.ansStep += temp.y + ‘0‘;
48                 que[tail++] = temp;
49             }
50         }
51     }
52 }
53
54 int main() {
55     head = 0, tail = 0;
56     memset(vis, 0, sizeof(vis));
57
58     for (int i=0; i<5; ++i) {
59         for (int j=0; j<5; ++j) {
60             cin >> mp[i][j];
61         }
62     }
63     now.x = 0, now.y = 0;
64     now.ansStep += "00";
65
66     vis[0][0] = 1;
67     que[tail++] = now;
68     bfs();
69
70     string ansstr = ans.ansStep;
71     int len = ansstr.length();
72     for (int i=0; i<len && i+2<len; i+=2) {
73         cout << "(" << ansstr[i] << ", " << ansstr[i+1] << ")" << endl;
74     }
75     return 0;
76 }