ICPC2018沈阳网络赛 Lattice's basics in digital electronics（模拟）

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Lattice‘s basics in digital electronics

44.08%
1000ms
131072K

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

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

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

For example, consider the given data "101010101010101010101010", it should be divided into

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 $S_iSi? to an ASCII code C_iCi?, and the message length MM, they want to peek his privacy so they come to you to write a program that decodes messages that LATTICE receives.$

Input

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

Every test case starts with one line containing two integers, $M\ (0<M\leq100000)M (0<M≤100000), the number of original characters, and N\ (1\leq N \leq 256)N (1≤N≤256), then NN lines, every line contains an integer C_iCi?, and a string S_i(0<|S_i|\leq 10)Si?(0<∣Si?∣≤10), means that S_iSi? represents C_iCi?, the ASCII code to a visible character and S_iSi? only contains ‘0‘or ‘1‘ and there are no two numbers ii and jj that S_iSi? is prefix of S_jSj?.$

Then one line contains data that is going to be received in hexadecimal.

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

Hint

Lattice‘s encoding rule for test case

ASCII code	character	lattice‘s code

the device takes this input in hex

14DB24722698

input in binary

 
0001 0100 1101 1011 0010 0100 0111 0010 0010 0110 1001 1000

formatted into

parity check of the third line and the last line failed, so ignore those two lines.parity bits should also be ignored.

arrange those bits by the rules informed

 
0001 01001 011 011 01001 0001 011 01001

output the result

12332132

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
8 3
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

大模拟，进制转换解码。900ms...多交几发就过了

#include<bits/stdc++.h>
#define MAX 15
using namespace std;
typedef long long ll;

int x;
char y[MAX];
string s,ss;
map<string,int> mp;
string bb(char x){
    if(x==‘0‘) return "0000";if(x==‘8‘) return "1000";
    if(x==‘1‘) return "0001";if(x==‘9‘) return "1001";
    if(x==‘2‘) return "0010";if(x==‘A‘||x==‘a‘) return "1010";
    if(x==‘3‘) return "0011";if(x==‘B‘||x==‘b‘) return "1011";
    if(x==‘4‘) return "0100";if(x==‘C‘||x==‘c‘) return "1100";
    if(x==‘5‘) return "0101";if(x==‘D‘||x==‘d‘) return "1101";
    if(x==‘6‘) return "0110";if(x==‘E‘||x==‘e‘) return "1110";
    if(x==‘7‘) return "0111";if(x==‘F‘||x==‘f‘) return "1111";
}
int find(string s){
    int c=0;
    for(int i=0;i<s.length();i++){
        if(s[i]==‘1‘) c++;
    }
    return c;
}
int main()
{
    int t,n,m,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        mp.clear();
        for(i=1;i<=m;i++){
            scanf("%d %s",&x,y);
            mp[y]=x;
        }
        cin>>s;
        ss="";
        int len=s.length();
        for(i=0;i<len;i++){
            ss+=bb(s[i]);
        }
        s="";
        len=ss.length();
        for(i=0;i<len;i+=9){
            if(i+8>=len) break;
            if((find(ss.substr(i,8))&1)!=ss[i+8]-‘0‘) s+=ss.substr(i,8);
        }
        len=s.length();
        for(i=0;i<len;i++){
            if(n<=0) break;
            for(j=i;j<len;j++){
                if(mp[s.substr(i,j-i+1)]){
                    printf("%c",mp[s.substr(i,j-i+1)]);
                    n--;i=j;
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}