Wildcard
Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on HDU. Original ID: 390164-bit integer IO format: %I64d Java class name: Main
When specifying file names (or paths) in DOS, Microsoft Windows and Unix-like operating systems, the asterisk character (“*") substitutes for any zero or more characters, and thequestion mark (“?") substitutes for any one character.
Now give you a text and a pattern, you should judge whether the pattern matches the text or not.
Now give you a text and a pattern, you should judge whether the pattern matches the text or not.
Input
There are several cases. For each case, only two lines. The first line contains a text, which contains only lower letters. The last line contains a pattern, which consists of lower letters and the two wildcards (“*", "?").
The text is a non-empty string and its size is less than 100,000, so do the pattern.
We ensure the number of “?” and the number of “*” in the pattern are both no more than 10.
The text is a non-empty string and its size is less than 100,000, so do the pattern.
We ensure the number of “?” and the number of “*” in the pattern are both no more than 10.
Output
Output “YES” if the pattern matches the text, otherwise “NO”.
Sample Input
abcdef a*c*f
Sample Output
YES
Source
解题：AC自动机
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 200005; 4 vector<int>pos[maxn]; 5 int node[maxn][26],fail[maxn],tot,cnt; 6 char wild[maxn],str[maxn],tmp[maxn]; 7 void build(char *s) { 8 int root = tot = cnt = 0; 9 memset(node[0],0,sizeof node[0]); 10 pos[0].clear(); 11 for(int i = 0; !i || s[i-1]; ++i) { 12 if(s[i] == '?' || s[i] == 0) { 13 if(root) { 14 pos[root].push_back(i-1); 15 root = 0; 16 ++cnt; 17 } 18 continue; 19 } 20 int c = s[i] - 'a'; 21 if(!node[root][c]) { 22 node[root][c] = ++tot; 23 memset(node[tot],0,sizeof node[tot]); 24 pos[tot].clear(); 25 } 26 root = node[root][c]; 27 } 28 } 29 queue<int>q; 30 void getFail() { 31 while(!q.empty()) q.pop(); 32 for(int i = 0; i < 26; ++i) 33 if(node[0][i]) { 34 q.push(node[0][i]); 35 fail[node[0][i]] = 0; 36 } 37 while(!q.empty()) { 38 int u = q.front(); 39 q.pop(); 40 for(int i = 0; i < 26; ++i) { 41 if(node[u][i]) { 42 q.push(node[u][i]); 43 int v = node[u][i]; 44 fail[v] = node[fail[u]][i]; 45 if(pos[fail[v]].size()) 46 pos[v].insert(pos[v].end(),pos[fail[v]].begin(),pos[fail[v]].end()); 47 } else node[u][i] = node[fail[u]][i]; 48 } 49 } 50 } 51 int x[maxn]; 52 int Find(char *s,int root = 0) { 53 if(!cnt) return 0; 54 for(int i = 0; s[i]; ++i) { 55 root = node[root][s[i] - 'a']; 56 x[i] = 0; 57 for(int j = 0; j < pos[root].size(); ++j) 58 if(i >= pos[root][j]) { 59 ++x[i - pos[root][j]]; 60 if(x[i - pos[root][j]] == cnt) 61 return i - pos[root][j]; 62 } 63 } 64 return -1; 65 } 66 bool match(int slen,int plen) { 67 int letter = 0,s = -1; 68 for(int i = 0; i < plen; ++i) 69 letter += (wild[i] != '*'); 70 if(letter > slen) return false; 71 for(int i = 0; str[i]; ++i) { 72 if(wild[i] == '?' || str[i] == wild[i]) continue; 73 if(wild[i] == '*') s = i + 1; 74 else return false; 75 break; 76 } 77 if(s == -1) return true; 78 for(int i = slen-1,j = plen-1; ; --i,--j) { 79 if(i == -1 && j == -1) break; 80 if(i == -1) { 81 if(wild[j] != '*') return false; 82 break; 83 } 84 if(j == -1) return false; 85 if(wild[j] == '?' || str[i] == wild[j]) 86 wild[j] = str[slen = i] = 0; 87 else if(wild[j] == '*') break; 88 else return false; 89 } 90 91 int st = s - 1; 92 for(int i = 1; i < plen - letter; ++i) { 93 if(wild[s] == '*') { 94 ++s; 95 continue; 96 } 97 int len = 0; 98 while(wild[s] != '*' && wild[s]) { 99 tmp[len++] = wild[s++]; 100 tmp[len] = 0; 101 } 102 build(tmp); 103 getFail(); 104 int ps = Find(st + str); 105 if(ps == -1) return false; 106 st += ps + len; 107 if(st > slen) return false; 108 } 109 return true; 110 } 111 int main() { 112 while(~scanf("%s%s",str,wild)) { 113 bool flag = match(strlen(str),strlen(wild)); 114 puts(flag?"YES":"NO"); 115 } 116 return 0; 117 }