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 }
