暴力/set Codeforces Round #291 (Div. 2) C. Watto and Mechanism

 

题目传送门

 1 /*  2  set的二分查找  3  如果数据规模小的话可以用O(n^2)的暴力想法  4  否则就只好一个一个的换(a, b, c)，在set容器找相匹配的  5 */  6 #include <cstdio>  7 #include <cmath>  8 #include <string>  9 #include <cstring>  10 #include <iostream>  11 #include <algorithm>  12 #include <map>  13 #include <set>  14 #include <vector>  15 using namespace std;  16  17 int main(void)  18 {  19 //freopen ("C.in", "r", stdin);  20  21 set<string> s;  22 string tmp;  23 int n, m, mx;  24  25 while (cin >> n >> m)  26  {  27 mx = -1;  28  s.clear ();  29 for (int i=1; i<=n; ++i)  30  {  31 cin >> tmp;  32  s.insert (tmp);  33 int len = tmp.size ();  34 mx = max (mx, len);  35  }  36 if (n * m * mx < 1e8)  37  {  38 for (int j=1; j<=m; ++j)  39  {  40 cin >> tmp;  41 int cnt = 0;  42 set<string>::iterator it;  43 for (it=s.begin (); it!=s.end (); ++it)  44  {  45 if (it->size () == tmp.size ())  46  {  47 cnt = 0;  48 for (int j=0; j<it->size (); ++j)  49  {  50 if ((*it)[j] != tmp[j]) cnt++;  51 if (cnt > 1) break;  52  }  53 if (cnt == 1) break;  54  }  55  }  56 if (cnt == 1) cout << "YES" << endl;  57 else cout << "NO" << endl;  58  }  59 continue;  60  }  61  62 for (int i=1; i<=m; ++i)  63  {  64 cin >> tmp;  65 bool flag = false;  66 for (int j=0; tmp[j]!='\0'; ++j)  67  {  68 if (tmp[j] == 'a')  69  {  70 tmp[j] = 'b';  71 if (s.find (tmp) != s.end ())  72  {  73 flag = true; break;  74  }  75 tmp[j] = 'c';  76 if (s.find (tmp) != s.end ())  77  {  78 flag = true; break;  79  }  80 tmp[j] = 'a';  81  }  82 if (tmp[j] == 'b')  83  {  84 tmp[j] = 'a';  85 if (s.find (tmp) != s.end ())  86  {  87 flag = true; break;  88  }  89 tmp[j] = 'c';  90 if (s.find (tmp) != s.end ())  91  {  92 flag = true; break;  93  }  94 tmp[j] = 'b';  95  }  96 if (tmp[j] == 'c')  97  {  98 tmp[j] = 'a';  99 if (s.find (tmp) != s.end ()) 100  { 101 flag = true; break; 102  } 103 tmp[j] = 'b'; 104 if (s.find (tmp) != s.end ()) 105  { 106 flag = true; break; 107  } 108 tmp[j] = 'c'; 109  } 110  } 111 (flag) ? cout << "YES" << endl : cout << "NO" << endl; 112  } 113  } 114 115 return 0; 116 } 117 118 /* 119 YES NO 120 */