DFS/BFS Codeforces Round #301 (Div. 2) C. Ice Cave

 

题目传送门

 1 /*  2  题意：告诉起点终点，踩一次, '.'变成'X'，再踩一次，冰块破碎，问是否能使终点冰破碎  3  DFS：如题解所说，分三种情况：1. 如果两点重合，只要往外走一步再走回来就行了；2. 若两点相邻，  4  那么要不就是踩一脚就破了或者踩一脚走开再走回来踩一脚破了；3. 普通的搜索看是否能到达，  5  若能还是要讨论终点踩几脚的问题:)  6  DFS 耗时大，险些超时，可以用BFS来做  7 */  8 #include <cstdio>  9 #include <algorithm> 10 #include <cstring> 11 #include <string> 12 #include <iostream> 13 using namespace std; 14 15 const int MAXN = 5e2 + 10; 16 const int INF = 0x3f3f3f3f; 17 int n, m; 18 int sx, sy, ex, ey; 19 char maze[MAXN][MAXN]; 20 bool vis[MAXN][MAXN]; 21 int dx[4] = {1, -1, 0, 0}; 22 int dy[4] = {0, 0, -1, 1}; 23 24 void DFS(int x, int y) 25 { 26 vis[x][y] = true; 27 if (maze[x][y] == 'X') return ; 28 29 for (int i=0; i<4; ++i) 30  { 31 int tx = x + dx[i]; 32 int ty = y + dy[i]; 33 34 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && !vis[tx][ty]) 35  { 36  DFS (tx, ty); 37  } 38  } 39 40 return ; 41 } 42 43 int main(void) //Codeforces Round #301 (Div. 2) C. Ice Cave 44 { 45 //freopen ("C.in", "r", stdin); 46 47 while (scanf ("%d%d", &n, &m) == 2) 48  { 49 memset (vis, 0, sizeof (vis)); 50 for (int i=1; i<=n; ++i) 51  { 52 scanf ("%s", maze[i]+1); 53  } 54 scanf ("%d%d", &sx, &sy); 55 scanf ("%d%d", &ex, &ey); 56 57 int cnt = 0; bool flag = false; 58 for (int i=0; i<4; ++i) 59  { 60 int tx = ex + dx[i]; 61 int ty = ey + dy[i]; 62 if (tx == sx && ty == sy) flag = true; 63 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.') cnt++; 64  } 65 66 if (sx == ex && sy == ey) 67  { 68 if (cnt >= 1) puts ("YES"); 69 else puts ("NO"); 70  } 71 else if (flag) 72  { 73 if (maze[ex][ey] == 'X') puts ("YES"); 74 else 75  { 76 if (cnt >= 1) puts ("YES"); 77 else puts ("NO"); 78  } 79  } 80 else 81  { 82 maze[sx][sy] = '.'; 83  DFS (sx, sy); 84 if (vis[ex][ey]) 85  { 86 if (maze[ex][ey] == 'X') puts ("YES"); 87 else if (cnt >= 2) puts ("YES"); 88 else puts ("NO"); 89  } 90 else puts ("NO"); 91  } 92  } 93 94 return 0; 95 }

 1 #include <cstdio>

 2 #include <algorithm>

 3 #include <cstring>

 4 #include <queue>

 5 #include <string>

 6 #include <iostream>

 7 using namespace std;  8 

 9 const int MAXN = 5e2 + 10; 10 const int INF = 0x3f3f3f3f; 11 int n, m; 12 int sx, sy, ex, ey; 13 char maze[MAXN][MAXN]; 14 bool vis[MAXN][MAXN]; 15 int dx[4] = {1, -1, 0, 0}; 16 int dy[4] = {0, 0, -1, 1}; 17 

18 bool BFS(void) 19 { 20     queue<pair<int, int> > Q; 21  Q.push (make_pair (sx, sy)); 22 

23     while (!Q.empty ()) 24  { 25         int x = Q.front ().first;    int y = Q.front ().second; Q.pop (); 26         for (int i=0; i<4; ++i) 27  { 28             int tx = x + dx[i]; 29             int ty = y + dy[i]; 30 

31             if (tx == ex && ty == ey)    return true; 32 

33             if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.') 34  { 35                 maze[tx][ty] = 'X'; 36  Q.push (make_pair (tx, ty)); 37  } 38  } 39  } 40 

41     return false; 42 } 43 

44 int main(void)        //Codeforces Round #301 (Div. 2) C. Ice Cave

45 { 46     //freopen ("C.in", "r", stdin);

47 

48     while (scanf ("%d%d", &n, &m) == 2) 49  { 50         memset (vis, 0, sizeof (vis)); 51         for (int i=1; i<=n; ++i) 52  { 53             scanf ("%s", maze[i]+1); 54  } 55         scanf ("%d%d", &sx, &sy); 56         scanf ("%d%d", &ex, &ey); 57 

58         int cnt = 0;    bool flag = false; 59         for (int i=0; i<4; ++i) 60  { 61             int tx = ex + dx[i]; 62             int ty = ey + dy[i]; 63             if (tx == sx && ty == sy)    flag = true; 64             if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.')    cnt++; 65  } 66 

67         if (sx == ex && sy == ey) 68  { 69             if (cnt >= 1)    puts ("YES"); 70             else puts ("NO"); 71  } 72         else if (flag) 73  { 74             if (maze[ex][ey] == 'X')    puts ("YES"); 75             else

76  { 77                 if (cnt >= 1)    puts ("YES"); 78                 else    puts ("NO"); 79  } 80  } 81         else

82  { 83             maze[sx][sy] = '.'; 84             //DFS (sx, sy);

85             if (BFS () == true) 86  { 87                 if (maze[ex][ey] == 'X')    puts ("YES"); 88                 else if (cnt >= 2)    puts ("YES"); 89                 else    puts ("NO"); 90  } 91             else    puts ("NO"); 92  } 93  } 94 

95     return 0; 96 }

BFS做法