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。。
1 #include<algorithm>
2 #include<iostream>
3 #include<cstdlib>
4 #include<cstring>
5 #include<cstdio>
6 #include<vector>
7 using std::vector;
8 const int Max_N = 60010;
9 struct Node {
10 int idx;
11 Node *next[10];
12 void set() {
13 idx = 0;
14 for (int i = 0; i < 10; i++) next[i] = NULL;
15 }
16 };
17 struct Trie {
18 int l;
19 char text[Max_N], buf[64];
20 Node stack[Max_N * 10], *root, *tail;
21 inline void link(char *src) {
22 for (int j = 0; src[j] != ‘\0‘; j++) text[l++] = src[j];
23 text[l] = ‘\0‘;
24 }
25 void init(){
26 l = 0;
27 tail = &stack[0];
28 root = tail++;
29 root->set();
30 }
31 inline Node *newNode(){
32 Node *p = tail++;
33 p->set();
34 return p;
35 }
36 inline void insert(Node *x, char *src, int idx){
37 char *p = src;
38 while (*p != ‘\0‘){
39 if (!x->next[*p - ‘0‘]) x->next[*p - ‘0‘] = newNode();
40 x = x->next[*p - ‘0‘];
41 p++;
42 }
43 x->idx = idx;
44 }
45 inline int query(Node *x, char *src){
46 char *p = src;
47 while (*p != ‘\0‘){
48 if (x->idx) return x->idx;
49 if (!x || !x->next[*p - ‘0‘]) return -1;
50 x = x->next[*p - ‘0‘];
51 p++;
52 }
53 return -1;
54 }
55 inline void insert(char *src, int idx){
56 insert(root, src, idx);
57 }
58 inline int query(char *src){
59 return query(root, src);
60 }
61 inline void gogo() {
62 vector<int> res;
63 for (int i = 0; text[i] != ‘\0‘; i++) {
64 int ret = query(text + i);
65 if (ret != -1) res.push_back(ret);
66 }
67 int n = res.size();
68 if (n) {
69 printf("Found key: ");
70 for (int i = 0; i < n; i++) {
71 printf("[Key No. %d]%c", res[i], i < n - 1 ? ‘ ‘ : ‘\n‘);
72 }
73 } else {
74 puts("No key can be found !");
75 }
76 }
77 }solve;
78 int main() {
79 #ifdef LOCAL
80 freopen("in.txt", "r", stdin);
81 freopen("out.txt", "w+", stdout);
82 #endif
83 int n, m;
84 char buf[64];
85 while (~scanf("%d %d", &n, &m)) {
86 solve.init();
87 while (n--) {
88 scanf("%s", buf);
89 solve.link(buf);
90 }
91 getchar(); getchar();
92 for (int i = 1; i <= m; i++) {
93 gets(buf);
94 solve.insert(strchr(buf, ‘]‘) + 2, i);
95 }
96 solve.gogo();
97 }
98 return 0;
99 }