Problem Description
3 AA BB CC ooxxCC%dAAAoen....END
AA: 2 CC: 1HintHit: 题目描述中没有被提及的所有情况都应该进行考虑。比如两个病毒特征码可能有相互包含或者有重叠的特征码段。 计数策略也可一定程度上从Sample中推测。
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; int const MAX = 2000005; char word[1005][55], text[MAX]; int cnt[1005]; struct node { int id; bool end; node *next[26]; node *fail; node() { id = -1; end = false; memset(next, NULL, sizeof(next)); fail = NULL; } }; void Insert(node *p, char *s, int id) { for(int i = 0; s[i] != '\0'; i++) { int idx = s[i] - 'A'; if(p -> next[idx] == NULL) p -> next[idx] = new node(); p = p -> next[idx]; } p -> end = true; p -> id = id; } void AC_Automation(node *root) { queue <node*> q; q.push(root); while(!q.empty()) { node *p = q.front(); q.pop(); for(int i = 0; i < 26; i++) { if(p -> next[i]) { if(p == root) p -> next[i] -> fail = root; else p -> next[i] -> fail = p -> fail -> next[i]; q.push(p -> next[i]); } else { if(p == root) p -> next[i] = root; else p -> next[i] = p -> fail -> next[i]; } } } } bool Query(node *root) { bool flag = false; int len = strlen(text); node *p = root; for(int i = 0; i < len; i++) { int idx = text[i] - 'A'; if(idx < 0 || idx > 25) { p = root; continue; } while(!p -> next[idx] && p != root) p = p -> fail; p = p -> next[idx]; if(!p) { p = root; continue; } node *tmp = p; while(tmp != root) { if(tmp -> end) { flag = true; cnt[tmp -> id]++; } else break; tmp = tmp -> fail; } } return flag; } int main() { int n; while(scanf("%d", &n) != EOF) { memset(cnt, 0, sizeof(cnt)); node *root = new node(); for(int i = 0; i < n; i++) { scanf("%s", word[i]); Insert(root, word[i], i); } AC_Automation(root); getchar(); gets(text); if(Query(root)) for(int i = 0; i < n; i++) if(cnt[i]) printf("%s: %d\n", word[i], cnt[i]); } }
原文地址:http://blog.csdn.net/tc_to_top/article/details/44103339