hdu 1277 全文检索 (直接映射查找 || 自动机)

Problem - 1277

　　无聊做水题的时候发现的一道题目。这道题第一反应可以用自动机来解决。当然，条件是各种限制，从而导致可以用直接映射标记的方法来搜索。具体的做法就像RK算法一样，将字符串hash成一个数，这里每一个关键字前四位都是不同的，这样就有利于hash搜索了。当前四位匹配的时候，就可以搜索整个串是否完全匹配。这整个的复杂度大概是O(n*m)，n是全文测长度，m是关键字的长度。

自动机代码：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <set>

 using namespace std;

 const int kind = ;
 struct Node {
     Node *next[kind];
     Node *fail;
     string label;
     Node() {
         memset(next, , sizeof(next));
         fail = NULL;
         label = "";
     }
     Node(char *tmp) {
         memset(next, , sizeof(next));
         fail = NULL;
         label = tmp;
     }
 } ;

 struct Trie {
     Node *Root;
     Trie() { Root = new Node();}

     void insert(char *s, char *label) {
         Node *p = Root;
         while (*s) {
             int idx = *s - '';
             if (!p->next[idx]) {
                 p->next[idx] = new Node();
                 p->next[idx]->fail = p;
             }
             p = p->next[idx];
             s++;
         }
         p->label = label;
     }
     void build() {
         queue<Node *> Q;
         while (!Q.empty()) Q.pop();
         Q.push(Root);
         Root->fail = NULL;
         Node *cur, *p;
         while (!Q.empty()) {
             cur = Q.front();
             Q.pop();
             for (int i = ; i < kind; i++) {
 //                cout << cur->next[i] << ' ';
                 if (cur->next[i]) {
 //                    cout << i << ' ';
                     Q.push(cur->next[i]);
                     p = cur->fail;
                     while (p) {
                         if (p->next[i]) {
                             cur->next[i]->fail = p->next[i];
                             break;
                         }
                         p = p->fail;
                     }
                     if (!p) cur->next[i]->fail = Root;
                 }
             }
         }
     }
     void query(char *s) {
         set<string> vis;
         bool pnt = false;
         Node *cur = Root;
         while (*s) {
             int idx = *(s++) - '';
             while (cur && !cur->next[idx]) cur = cur->fail;
             cur = cur ? cur->next[idx] : Root;
             if (vis.find(cur->label) != vis.end()) continue;
             if (cur->label.length()) {
                 if (!pnt) {
                     printf("Found key:");
                     pnt = true;
                 }
                 cout << ' ' << cur->label;
                 vis.insert(cur->label);
             }
         }
         if (!pnt) puts("No key can be found !");
         cout << endl;
     }
 } trie;

 char str[], buf[];

 int main() {
 //    freopen("in", "r", stdin);
     int n, m;
     while (cin >> n >> m) {
         trie = Trie();
         gets(buf);
         str[] = ;
         while (n-- && gets(buf)) strcat(str, buf);
         gets(buf);
         while (m-- && gets(buf)) {
             char *p = buf;
             while (*(p++) != ']') ;
             *(p++) = ;
             while (*p == ' ') p++;
             trie.insert(p, buf);
 //            cout << p << ' ' << buf << " inserted!"<< endl;
         }
         trie.build();
 //        cout << "built!!" << endl;
         trie.query(str);
     }
     return ;
 }

UPD：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <string>

 using namespace std;

 const int HASHLEN = ;
 const int M = ;
 const int N = ;
 string label[N], key[N];
 bool vis[N];
 int id[M];
 char str[ * N], buf[N];

 bool check(int t, char *p) {
     int len = key[t].length();
     for (int i = ; i < len; i++) {
         if (key[t][i] != p[i]) return false;
     }
     return true;
 }

 void work(char *p) {
     bool pnt = false;
     char *q = p;
     int cur = , ep = ;
     for (int i = ; i < HASHLEN - ; i++) cur = cur *  + *(q++) - '', ep *= ;
     while (*q && *p) {
 //        cout << cur << endl;
         cur = cur *  + *q - '';
         cur %= ep;
         int t = id[cur];
         if (t != - && !vis[t]) {
             if (check(t, p)) {
                 if (!pnt) printf("Found key:");
                 pnt = vis[t] = true;
                 cout << ' ' << label[t];
             }
         }
         p++, q++;
     }
     if (!pnt) printf("No key can be found !");
     cout << endl;
 }

 int main() {
 //    freopen("in", "r", stdin);
     int n, m;
     while (cin >> n >> m) {
         memset(id, -, sizeof(id));
         memset(vis, , sizeof(vis));
         gets(buf);
         str[] = ;
         while (n-- && gets(buf)) strcat(str, buf);
         gets(buf);
         for (int i = ; i < m; i++) {
             gets(buf);
             char *p = buf;
             while (*p && *p != ']') p++;
             *(++p) = ;
             p++;
             while (*p && *p == ' ') p++;
             label[i] = buf;
             key[i] = p;
 //            cout << label[i] << ' ' << key[i] << endl;
             int cid = ;
             for (int i = ; i < HASHLEN; i++) cid = cid *  + p[i] - '';
 //            cout << cid << endl;
             if (id[cid] != -) { puts("shit!"); while () ;}
             id[cid] = i;
         }
         work(str);
     }
     return ;
 }

　　不知道为什么这个代码能过G++但是C++过不了，应该是代码里面有bug。另外，数据似乎跟描述的有点出入，说好的不同key值的前四位不会相同，可是写了句来试它的数据时就发现被骗了，所以后来用了5位，过了！

——written by Lyon