标签：poj cut null build ems 子串 之一 alien node
描述
Aliens on planet Pandora also write computer programs like us. Their programs only consist of capital letters (‘A’ to ‘Z’) which they learned from the Earth. On planet Pandora, hackers make computer virus, so they also have anti-virus software. Of course they learned virus scanning algorithm from the Earth. Every virus has a pattern string which consists of only capital letters. If a virus’s pattern string is a substring of a program, or the pattern string is a substring of the reverse of that program, they can say the program is infected by that virus. Give you a program and a list of virus pattern strings, please write a program to figure out how many viruses the program is infected by.输入There are multiple test cases. The first line in the input is an integer T ( T <= 10) indicating the number of test cases.
For each test case:
The first line is a integer n( 0 < n <= 250) indicating the number of virus pattern strings.
Then n lines follows, each represents a virus pattern string. Every pattern string stands for a virus. It’s guaranteed that those n pattern strings are all different so there are n different viruses. The length of pattern string is no more than 1,000 and a pattern string at least consists of one letter.
The last line of a test case is the program. The program may be described in a compressed format. A compressed program consists of capital letters and “compressors”. A “compressor” is in the following format:
[qx]
q is a number( 0 < q <= 5,000,000)and x is a capital letter. It means q consecutive letter xs in the original uncompressed program. For example, [6K] means ‘KKKKKK’ in the original program. So, if a compressed program is like:
AB[2D]E[7K]G
It actually is ABDDEKKKKKKKG after decompressed to original format.
The length of the program is at least 1 and at most 5,100,000, no matter in the compressed format or after it is decompressed to original format.输出For each test case, print an integer K in a line meaning that the program is infected by K viruses.
样例输入
3
2
AB
DCB
DACB
3
ABC
CDE
GHI
ABCCDEFIHG
4
ABB
ACDEE
BBB
FEEE
A[2B]CD[4E]F
样例输出
0
3
2
提示
In the second case in the sample input, the reverse of the program is ‘GHIFEDCCBA’, and ‘GHI’ is a substring of the reverse, so the program is infected by virus ‘GHI’.
1 #include <iostream> 2 #include <string.h> 3 #include <algorithm> 4 #include <stack> 5 #include <string> 6 #include <math.h> 7 #include <queue> 8 #include <stdio.h> 9 #include <string.h> 10 #include <vector> 11 #include <fstream> 12 #include <set> 13 14 using namespace std; 15 const int maxn = 255; 16 char line[5100005]; 17 int kase, n, infecsum = 0, nodecou,l; 18 struct node { 19 node*next[26]; 20 node*prev; 21 int count; 22 node() { 23 memset(next, 0, sizeof(next)); 24 prev = NULL; 25 count = 0; 26 } 27 }tree[maxn * 1000]; 28 29 void build() { 30 for (int i = 0; i < 26; i++) 31 tree[0].next[i] = tree + 1; 32 tree[0].prev = NULL; 33 tree[1].prev = tree; 34 queue<node*>q; 35 q.push(tree + 1); 36 while (!q.empty()) { 37 node*now = q.front(); 38 q.pop(); 39 for (int i = 0; i < 26; i++) { 40 node*child = now->next[i]; 41 if (child) { 42 node*prev = now->prev; 43 while (prev->next[i] == NULL) 44 prev = prev->prev; 45 child->prev = prev->next[i]; 46 q.push(child); 47 } 48 } 49 } 50 } 51 52 int search(char *str) { 53 int ans = 0; 54 node*first = tree + 1; 55 for (int i = 0; str[i] != ‘\0‘; i++) { 56 while (first->next[str[i] - ‘A‘] == NULL&&first!=(tree+1)) 57 first = first->prev; 58 if (first->next[str[i] - ‘A‘]) 59 first = first->next[str[i] - ‘A‘]; 60 node*tmp = first; 61 while (tmp != NULL && tmp->count != -1) { 62 ans += tmp->count; 63 tmp->count = -1; 64 tmp = tmp->prev; 65 } 66 } 67 return ans; 68 } 69 70 void insert(node*rt, char* line) { 71 int l = strlen(line); 72 for (int i = 0; i < l; i++) { 73 if (rt->next[line[i] - ‘A‘] == NULL) { 74 nodecou++; 75 rt->next[line[i] - ‘A‘] = tree + nodecou; 76 } 77 rt = rt->next[line[i] - ‘A‘]; 78 } 79 rt->count++; 80 } 81 82 void stringin() { 83 char ch; 84 scanf("\n"); 85 while (scanf("%c", &ch)) { 86 if (ch == ‘\n‘) 87 break; 88 if (ch == ‘[‘) { 89 int c; 90 scanf("%d", &c); 91 scanf("%c", &ch); 92 for (int i = 1; i <= c; i++)line[l++] = ch; 93 scanf("%c", &ch); 94 } 95 else 96 line[l++] = ch; 97 } 98 line[l] = ‘\0‘; 99 } 100 101 void init() { 102 scanf("%d", &kase); 103 while (kase--) { 104 memset(tree, 0, sizeof(tree)); 105 infecsum = 0, nodecou = 1,l=0; 106 scanf("%d", &n); 107 for (int i = 1; i <= n; i++) { 108 scanf("%s",line); 109 insert(tree + 1, line); 110 } 111 build(); 112 stringin(); 113 infecsum=search(line); 114 reverse(line,line+l); 115 infecsum+=search(line); 116 printf("%d\n", infecsum); 117 } 118 } 119 120 int main() 121 { 122 init(); 123 return 0; 124 }
这题杀我……！
果然自动AC什么的都是假的QwQ
一开始倔强地使用了string结果无限TLE
后来又因为中间少了一句迭代而WA
die了
再也不想看到这道题
开学来做得最悲伤的一道题没有之一
主要是要考虑子串中套子串的问题如果不加标记会TLE（我看见有人没有考虑AC了？？逗我？？）
稍微处理一下字符串
18.10.29 POJ 3987 Computer Virus on Planet Pandora（AC自动机+字符串处理）
标签：poj cut null build ems 子串 之一 alien node
原文地址：https://www.cnblogs.com/yalphait/p/9874022.html
