Description

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set
of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols {A, B, C, D}

The following code is immediately decodable:

A:01
B:10 C:0010 D:0000

but this one is not:

A:01
B:10 C:010 D:0000
 (Note that A is
a prefix of C)

Input

Write a program that accepts as input a series of groups of records from a data file. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a
single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

Output

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group
is, or is not, immediately decodable.

The Sample Input describes the examples above.

Sample Input

01
10
0010
0000
9
01
10
010
0000
9

Sample Output

Set 1 is immediately decodable
Set 2 is not immediately decodable

HINT

//(如果任意一个串,均不是其它串的前缀则表明,is immediately decodable,否则is not immediately decodable)

#include<stdio.h>
#include<string.h>
int main()
{
char a[100][100];
int t=0,i,j,t1,t2,k,s=1;
while(gets(a[t]))
{
t++;
while(gets(a[t]))
{
if(a[t][0]=='9')
break;
t++;
}
for(i=0;i<t;i++)
{
t1=strlen(a[i]);
for(j=i+1;j<t;j++)
{
//t2=strlen(a[j]);
for(k=0;k<t1;k++)
{
if(a[i][k]!=a[j][k])//(is immediately decodable)
break;
}
if(k==t1)//(is not immediately decodable)
break;
}
if(j!=t)//(is not immediately decodable)
break;
}
if(i==t)
printf("Set %d is immediately decodable\n",s);
else
printf("Set %d is not immediately decodable\n",s); //for(i=0;i<t;i++)
//printf("%s\n",a[i]);
//printf("\n");
t=0;
s++;
}
return 0;
}

//参考他人代码

单模式匹配算法:给定一个单词和一个字符串,查看字符串中是否存在该单词,通过调用strstr函数进行匹配;

#include<iostream>
#include<string>
#include<stdio.h>
#include<string.h> using namespace std ; int main()
{
char word[1000][20] ;
int t = 1 ;
while(cin >> word[0])
{
int i = 1 ;
while(cin >> word[i++])
if(strcmp(word[i-1],"9")==0)
break ;
bool flag = true ;
i--;
for(int j = 0 ; j < i ; j++)//每一行均与其它进行比较
{
char *p = NULL ;
for(int k = 0 ; k < i ; k++)
{
if(j == k)
continue ;
p = strstr(word[j],word[k]) ;//第j行在其它行(k)中遍历,看是否是其前缀。
if(p == word[j])
{
flag = false ;
break ;
}
}
if(flag == false)
break ;
}
if(flag)
printf("Set %d is immediately decodable\n",t++) ;
else
printf("Set %d is not immediately decodable\n",t++) ;
}
return 0 ;
}

Immediate Decodability的更多相关文章

  1. UVa 644 Immediate Decodability

    吐槽下我的渣渣英语啊,即使叫谷歌翻译也没有看懂,最后还是自己读了好几遍题才读懂. 题目大意:题意很简单,就是给一些互不相同的由'0','1'组成的字符串,看看有没有一个字符串是否会成为另一个的开头的子 ...

  2. hdu 1305 Immediate Decodability(字典树)

    Immediate Decodability Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/O ...

  3. Immediate Decodability(字典树)

    Immediate Decodability Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/O ...

  4. UVA644 Immediate Decodability

    UVA644 Immediate Decodability Trie Trie模板题 难度几乎相等的题P2580 于是他错误的点名开始了 对于每组数据都清空树太浪费时间,所以我们只要在需要新点时预先把 ...

  5. POJ1056 IMMEDIATE DECODABILITY & POJ3630 Phone List

    题目来源:http://poj.org/problem?id=1056   http://poj.org/problem?id=3630 两题非常类似,所以在这里一并做了. 1056题目大意: 如果一 ...

  6. HDU 1305 Immediate Decodability 可直接解码吗?

    题意:一个码如果是另一个码的前缀,则 is not immediately decodable,不可直接解码,也就是给一串二进制数字给你,你不能对其解码,因解码出来可能有多种情况. 思路:将每个码按长 ...

  7. 「UVA644」 Immediate Decodability(Trie

    题意翻译 本题有多组数据.每组数据给出一列以"9"结尾的仅包含'0'和'1'的字符串,如果里面有一个是另一个的子串,输出"Set &case is not imm ...

  8. POJ 1056 IMMEDIATE DECODABILITY

    IMMEDIATE DECODABILITY Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9630   Accepted: ...

  9. HDU1305 Immediate Decodability (字典树

    Immediate Decodability An encoding of a set of symbols is said to be immediately decodable if no cod ...

  10. hdu 1305 Immediate Decodability

    原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1305 字典树裸题,如下: #include<algorithm> #include< ...

随机推荐

  1. 快速傅里叶变换FFT

    多项式乘法 #include <cstdio> #include <cmath> #include <algorithm> #include <cstdlib ...

  2. 浅析指针(pointer)与引用(reference)

    在c++函数中,形式参数用引用和用指针都可以起到在被调用函数中改变调用函数的变量的作用.什么时候用引用作参数?什么时候用指针作参数呢 void function (int *ptr); void fu ...

  3. 10.30 NFLS-NOIP模拟赛 解题报告

    总结:今天去了NOIP模拟赛,其实是几道USACO的经典的题目,第一题和最后一题都有思路,第二题是我一开始写了个spfa,写了一半中途发现应该是矩阵乘法,然后没做完,然后就没有然后了!第二题的暴力都没 ...

  4. express中路由设置的坑-----1

    router.get('/commodities/sortable', utils.logged, function (req, res) { Commodity.find({force_top:tr ...

  5. How Many Equations Can You Find(dfs)

    How Many Equations Can You Find Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 ...

  6. Orcale语句大全

    原文地址:http://www.cnblogs.com/omygod/archive/2007/08/31/876620.html Oracle 语句大全 1. Oracle安装完成后的初始口令?  ...

  7. java中解决request中文乱码问题

    request乱码问题(当我们提交的数据中含有中文信息时),分两种情况: 通过post方式提交数据给Servlet Servlet服务端部分代码: public void doPost(httpSer ...

  8. 项目适配iOS9遇到的一些问题及解决办法

    1.网络请求报错.升级Xcode 7.0发现网络访问失败.输出错误信息 The resource could not be loaded because the App Transport Secur ...

  9. Xcode工程使用CocoaPods管理第三方库新建工程时出现异常

    Xcode工程使用CocoaPods管理第三方库新建工程时出现异常 Xcode工程使用CocoaPods管理第三方库新建工程时出现错误工程使用CocoaPods管理第三方库,在新的目录update版本 ...

  10. ubuntu, Debian, CentOS

    ubuntu源自debian,内核很多文档都还是debian的字样,稳定性逐渐增强,基本满足日常开发. debian的核心稳定,性能强劲. centos的内核版本低,安全性高. 选择Debian是因为 ...