ACM-ICPC 2018 沈阳赛区网络预赛 I Lattice's basics in digital electronics [ 01字典树 ]

题目链接：

ACM-ICPC 2018 沈阳赛区网络预赛 I. Lattice's basics in digital electronics

题意概括：

题目给出一种处理字符串的编码规则：

• 以十六进制格式读入一个字符串，转换为二进制
• 把转换后得到的二进制字符串每 9 位划分为一个单元，模 9 后剩余的直接舍去
• 对于每个 9 位长的单元，对前 8 位进行奇偶校验：若第 9 位是 1，则若前 8 位的 1 个数是偶数时为真；第九位是 0，则若前 8 位的 1 个数是奇数时为真
• 校验为真则保留前 8 位，假就直接删去。校验后第 9 位一定删去
• 通过以上操作，得到了一个新的二进制字符串。题目会给一种特殊的编码方式，根据01编码得到一个 ASCII 码，输出便可
• 这种编码方式保证不会有包含性，也就是说某一个串不会是另一个串的前缀或后缀

数据范围：

 , 

 , 

题解分析：

根据十六进制串经过奇偶校验得到01串的过程就是单纯模拟，仔细一点就好

用得到的 01串 来匹配编码是这道题需要注意的地方，直接暴力肯定不现实

采用 01字典树 来存储所有的 01编码 ，在编码结束的节点存储编码对应的 ASCII 值，其余节点存 0

直接从 01串 头部开始在字典树上进行搜索，当搜索到某一节点值不为 0 时，便说明匹配到一个码

由于某一个串不会是另一个串的前缀或后缀，不需要考虑 fail指针。当匹配到一个码后，直接跳转回到根节点继续寻找下一个

AC代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

char trans[30][5] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111","1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111", "1010", "1011", "1100", "1101", "1110", "1111"};

char HTB[110][5], s[200010], bin[200010 * 4], data[200010 * 4];
int cnt = 0, M, N;

typedef struct Node {
    Node *next[2];
    int asc;
}trie;

trie *root;

void init(trie *p) {
    p->asc = 0;
    memset(p->next, NULL, sizeof(p->next));
}

void insert(char *number, int asc) {
    trie *p = root;
    for(int i = 0; number[i] != 0; i++) {
        int v = number[i] - 48;
        if(p->next[v] == NULL) {
            trie* temp = (trie*)malloc(sizeof(trie));
            init(temp);
            p->next[v] = temp;
        }
        p = p->next[v];
    }
    p->asc = asc;
}

void Clear(trie *p) {
    if (p == NULL) return;
    for (int i = 0; i < 2; i++) {
        if (p->next[i] != NULL) Clear(p->next[i]);
    }
    delete p;
}

void search(char s[]) {
    trie *p = root;
    for(int i = 0; s[i]; i++) {
        int v = s[i] - 48;
            p = p->next[v];
        if (p->asc) {
            if (cnt < M)
            printf("%c", p->asc);
            cnt ++;
            p = root;
        }
    }
}

bool check(char s[]) {
    int cnt = 0;
    for (int i = 0; i < 8; i++)
        if (s[i] == '1') cnt ++;
    if (s[8] == '1' && !(cnt % 2)) return true;
    if (s[8] == '0' && (cnt % 2)) return true;
    return false;
}

int main() {
    for (int i = 0; i <= 9; i ++)
        strcpy(HTB[48 + i], trans[i]);
    for (int i = 0; i < 6; i ++)
        strcpy(HTB[65 + i], trans[10 + i]);
    for (int i = 0; i < 6; i ++)
        strcpy(HTB[97 + i], trans[16 + i]);
    
    int T;
    scanf("%d", &T);
    while (T --) {
    scanf("%d%d", &M, &N);
    cnt = 0;
    root = (trie*)malloc(sizeof(trie));
    init(root);
    char S[20];
    int asc;
        
    for (int i = 0; i < N; i++) {
        scanf("%d%s", &asc, S);
        insert(S, asc);
    }
    
    scanf("%s", s);
    memset(bin, 0, sizeof(bin));
    memset(data, 0, sizeof(data));
    int ptr = 0;
    for (int i = 0; s[i] != 0; i ++) {
        for (int k = 0; k < 4; k++)
            bin[ptr++] = HTB[s[i]][k];
    }
    
    ptr = 0;
    for (int i = 0; bin[i] != 0; i += 9) {
        if (check(bin + i)) {
            for (int j = 0; j < 8; j++)
                data[ptr++] = bin[i + j];
        }
    }
    
    search(data);
    printf("\n");
    Clear(root);
    }
    return 0;
}

 

                                   Lattice's basics in digital electronics 

LATTICE is learning Digital Electronic Technology. He is talented, so he understood all those pieces of knowledge in 10−9 second. In the next 10−9 second, he built a data decoding device that decodes data encoded with his special binary coding rule to meaningful words.

His coding rule is called "prefix code", a type of code system (typically a variable-length code) distinguished by its possession of the "prefix property", which requires that there is no whole code word in the system that is a prefix (initial segment) of any other code word in the system. Note that his code is composed of only 0 and 1.

LATTICE's device only receives data that perfectly matches LATTICE's rules, in other words,people who send message to LATTICE will always obey his coding rule. However, in the process of receiving data, there are errors that cannot avoid, so LATTICE uses parity check to detect error bytes, after every 8-bit data there is 1 bit called parity bit, which should be '0'

if there are odd number of '1' s in the previous 8 bits and should be '1' if there are even number of '1' s. If the parity bit does not meet the fact, then the whole 9 bits (including the parity bit) should be considered as invalid data and ignored. Data without parity bit is also considered as invalid data. Parity bits will be deleted after the parity check.

For example, consider the given data "101010101010101010101010" , it should be divided into 3parts: "101010101" , "010101010" and "101010" . For the first part, there are 4 '1' s in the first8 bits, and parity bit is '1' , so this part passed the check. For the second part, there are 4

'1' s and parity bit is '0' , so this part failed the check. For the third part, it has less than 9bits so it contains no parity bit, so this part also failed the check. The data after parity check is "10101010" , which is the first 8 bits of first part.

Data passed the parity check will go into a process that decodes LATTICE's code. The process is described in the following example: consider a situation that, "010" represents

'A' and "1011" represents 'B' , if the data after parity check is "01010110101011010010" , it can be divided into "010" + "1011" + "010" + "1011" + "010" + "010" , which means "ABABAA" . LATTICE's device is so exquisite that it can decode all visible characters in the ASCII table .

LATTICE is famous for his Talk show, some reporters have sneaked into his mansion, they stole the data LATTICE to decode in hexadecimal, the coding rule consists of N pairs of corresponding relations from a bit string Si to an ASCII code Ci, and the message length M,they want to peek his privacy so they come to you to write a program that decodes messages that LATTICE receives.

Input

The first line an integer T (T < 35) represents the number of test cases.

Every test case starts with one line containing two integers, M (0 < M ≤ 100000), the number of original characters, and N (1 ≤ N ≤ 256), then N lines, every line contains an integer Ci, and a string Si(0 < ∣Si∣ ≤ 10), means that Si represents Ci, the ASCII code to a visible character and Si only contains '0' or '1' and there are no two numbers i and j thatSi isprefixofSj.

Then one line contains data that is going to be received in hexadecimal. (0 < ∣data∣< 200000).

Output

For each test case, output the decoded message in a new line, the length of the decoded message should be the same with the length of original characters, which means you can stop decoding having outputted M characters. Input guarantees that it will have no less thanM valid characters and all given ASCII codes represent visible characters.

样例输⼊

2
15 9
32 0100
33 11
100 1011
101 0110
104 1010
108 00
111 100
114 0111
119 0101 A6Fd021171c562Fde1 83
49 0001
50 01001
51 011 14DB24722698

样例输出

hello world!!!!
12332132

题⽬来源

ACM-ICPC 2018 沈阳赛区网络预赛