• 18.10.29 POJ 3987 Computer Virus on Planet Pandora(AC自动机+字符串处理)


    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] != ''; 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("
    ");
     85     while (scanf("%c", &ch)) {
     86         if (ch == '
    ')
     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] = '';
     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
    ", infecsum);
    117     }
    118 }
    119 
    120 int main()
    121 {
    122     init();
    123     return 0;
    124 }
    View Code

    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

    注定失败的战争,也要拼尽全力去打赢它; 就算输,也要输得足够漂亮。
  • 相关阅读:
    Python中获取字典中最值对应的键
    Python中if __name__ == "__main__": 的作用
    Python中将打印输出导向日志文件
    python函数参数前面单星号(*)和双星号(**)的区别
    Python中用datetime包进行对时间的一些操作
    python中scipy.misc.logsumexp函数的运用场景
    Python中在脚本中引用其他文件函数的方法
    KS-检验(Kolmogorov-Smirnov test) -- 检验数据是否符合某种分布
    R如何检验类别变量(nominal variable)与其他变量之间的相关性
    机器学习-贝叶斯算法
  • 原文地址:https://www.cnblogs.com/yalphait/p/9874022.html
Copyright © 2020-2023  润新知