水模拟题。
1 /* 1483 */ 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 6 #define MAXN 10005 7 8 typedef struct { 9 char s[26]; 10 int len; 11 } node_t; 12 13 node_t node[MAXN]; 14 char s[26]; 15 int len, v; 16 int n, m; 17 int flag; 18 19 int abs(int x) { 20 return x<0 ? -x:x; 21 } 22 23 int min(int a, int b) { 24 return a<b ? a:b; 25 } 26 27 int max(int a, int b) { 28 return a<b ? a:b; 29 } 30 31 void solve() { 32 int i, j, k, p, tmp; 33 int stack[26], top; 34 flag = 2; 35 36 // check if is correct 37 for (i=0; flag&&i<n; ++i) { 38 if (node[i].len==len) { 39 // swap 2 char or 1 char wrong or just the same 40 k = 0; 41 top = 0; 42 for (j=0; k<=2&&j<len; ++j) { 43 if (node[i].s[j] != s[j]) { 44 stack[top++] = j; 45 ++k; 46 } 47 } 48 if (k == 0) { 49 flag = 0; 50 v = i; 51 break; 52 } else if (k==1 && flag>1) { 53 flag = 1; 54 v = i; 55 } else if (k==2 && flag>1) { 56 if (node[i].s[stack[0]]==s[stack[1]] && node[i].s[stack[1]]==s[stack[0]]) { 57 flag = 1; 58 v = i; 59 } 60 } 61 } else if (flag>1 && node[i].len-len==1) { 62 // one letter is missing 63 bool st = true; 64 for (j=k=0; k<node[i].len; ++k) { 65 if (node[i].s[k] == s[j]) 66 ++j; 67 else if (st) 68 st = false; 69 else 70 break; 71 } 72 if (k >= node[i].len) { 73 flag = 1; 74 v = i; 75 } 76 } else if (flag>1 && len-node[i].len==1) { 77 // too much 78 bool st = true; 79 for (k=0,j=0; j<len; ++j) { 80 if (node[i].s[k] == s[j]) 81 ++k; 82 else if (st) 83 st = false; 84 else 85 break; 86 } 87 if (j>=len) { 88 flag = 1; 89 v = i; 90 } 91 } 92 } 93 } 94 95 int main() { 96 int i, j, k, tmp; 97 98 #ifndef ONLINE_JUDGE 99 freopen("data.in", "r", stdin); 100 #endif 101 102 while (scanf("%d", &n) != EOF) { 103 for (i=0; i<n; ++i) { 104 scanf("%s", node[i].s); 105 node[i].len = strlen(node[i].s); 106 } 107 scanf("%d", &m); 108 while (m--) { 109 scanf("%s", s); 110 len = strlen(s); 111 112 solve(); 113 if (flag == 2) 114 printf("%s is unknown ", s); 115 else if (flag == 0) 116 printf("%s is correct ", s); 117 else if (flag == 1) 118 printf("%s is a misspelling of %s ", s, node[v].s); 119 } 120 } 121 122 return 0; 123 }