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http://acm.hdu.edu.cn/showproblem.php?pid=1277
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。
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。
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 !。
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
Found key: [Key No. 9] [Key No. 5]
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 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4550517.html
