hdu 1277 全文检索

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1277

全文检索

Description

我们大家经常用google检索信息，但是检索信息的程序是很困难编写的；现在请你编写一个简单的全文检索程序。
问题的描述是这样的：给定一个信息流文件，信息完全有数字组成，数字个数不超过60000个，但也不少于60个；再给定一个关键字集合，其中关键字个数不超过10000个，每个关键字的信息数字不超过60个，但也不少于5个；两个不同的关键字的前4个数字是不相同的；由于流文件太长，已经把它分成多行；请你编写一个程序检索出有那些关键字在文件中出现过。

Input

第一行是两个整数M，N；M表示数字信息的行数，N表示关键字的个数；接着是M行信息数字，然后是一个空行；再接着是N行关键字；每个关键字的形式是：[Key No. 1] 84336606737854833158。

Output

输出只有一行，如果检索到有关键字出现，则依次输出，但不能重复，中间有空格，形式如：Found key: [Key No. 9] [Key No. 5]；如果没找到，则输出形如：No key can be found !。

Sample Input

20 10
646371829920732613433350295911348731863560763634906583816269
637943246892596447991938395877747771811648872332524287543417
420073458038799863383943942530626367011418831418830378814827
679789991249141417051280978492595526784382732523080941390128
848936060512743730770176538411912533308591624872304820548423
057714962038959390276719431970894771269272915078424294911604
285668850536322870175463184619212279227080486085232196545993
274120348544992476883699966392847818898765000210113407285843
826588950728649155284642040381621412034311030525211673826615
398392584951483398200573382259746978916038978673319211750951
759887080899375947416778162964542298155439321112519055818097
642777682095251801728347934613082147096788006630252328830397
651057159088107635467760822355648170303701893489665828841446
069075452303785944262412169703756833446978261465128188378490
310770144518810438159567647733036073099159346768788307780542
503526691711872185060586699672220882332373316019934540754940
773329948050821544112511169610221737386427076709247489217919
035158663949436676762790541915664544880091332011868983231199
331629190771638894322709719381139120258155869538381417179544
000361739177065479939154438487026200359760114591903421347697

[Key No. 1] 934134543994403697353070375063
[Key No. 2] 261985859328131064098820791211
[Key No. 3] 306654944587896551585198958148
[Key No. 4] 338705582224622197932744664740
[Key No. 5] 619212279227080486085232196545
[Key No. 6] 333721611669515948347341113196
[Key No. 7] 558413268297940936497001402385
[Key No. 8] 212078302886403292548019629313
[Key No. 9] 877747771811648872332524287543
[Key No. 10] 488616113330539801137218227609

Sample Output

Found key: [Key No. 9] [Key No. 5]

字典树简单题，开始用kmp写t了%>_<%，换了字典树过了。。

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<vector>
 using std::vector;
 const int Max_N = ;
 struct Node {
     int idx;
     Node *next[];
     void set() {
         idx = ;
         for (int i = ; i < ; i++) next[i] = NULL;
     }
 };
 struct Trie {
     int l;
     char text[Max_N], buf[];
     Node stack[Max_N * ], *root, *tail;
     inline void link(char *src) {
         for (int j = ; src[j] != '\0'; j++) text[l++] = src[j];
         text[l] = '\0';
     }
     void init(){
         l = ;
         tail = &stack[];
         root = tail++;
         root->set();
     }
     inline Node *newNode() {
         Node *p = tail++;
         p->set();
         return p;
     }
     inline void insert(Node *x, char *src, int idx) {
         char *p = src;
         while (*p != '\0') {
             if (!x->next[*p - '']) x->next[*p - ''] = newNode();
             x = x->next[*p - ''];
             p++;
         }
         x->idx = idx;
     }
     inline int query(Node *x, char *src) {
         char *p = src;
         while (*p != '\0') {
             if (x->idx) return x->idx;
             if (!x || !x->next[*p - '']) return -;
             x = x->next[*p - ''];
             p++;
         }
         return -;
     }
     inline void insert(char *src, int idx) {
         insert(root, src, idx);
     }
     inline int query(char *src) {
         return query(root, src);
     }
     inline  void gogo() {
         vector<int> res;
         for (int i = ; text[i] != '\0'; i++) {
             int ret = query(text + i);
             if (ret != -) res.push_back(ret);
         }
         int n = res.size();
         if (n) {
             printf("Found key: ");
             for (int i = ; i < n; i++) {
                 printf("[Key No. %d]%c", res[i], i < n -  ? ' ' : '\n');
             }
         } else {
             puts("No key can be found !");
         }
     }
 }solve;
 int main() {
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int n, m;
     char buf[];
     while (~scanf("%d %d", &n, &m)) {
         solve.init();
         while (n--) {
             scanf("%s", buf);
             solve.link(buf);
         }
         getchar(); getchar();
         for (int i = ; i <= m; i++) {
             gets(buf);
             solve.insert(strchr(buf, ']') + , i);
         }
         solve.gogo();
     }
     return ;
 }