UVA - 644 Immediate Decodability

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.

Note: 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

分析：

字符串匹配。

暴力可过。

代码：

#include<stdio.h>

#include<string.h>

char str[100][100];

int com(int t,int x)

{

int n,m,i;

n=strlen(str[t]);

m=strlen(str[x]);

n=(n>m)?m:n;

for(i=0;i<n;i++)

{

if(str[t][i]!=str[x][i])

return 1;

}

return 0;

}

int main()

{

int t,leap,i;

t=0;

leap=0;

int cas=0;

while(gets(str[t])!=NULL)

{

if(str[t][0]=='9')

{

cas++;

if(leap)

printf("Set %d is not immediately decodable\n",cas);

else

printf("Set %d is immediately decodable\n",cas);

t=0;

leap=0;

continue;

}

for(i=0;i<t;i++)

{

if(com(t,i)==0)

{

leap=1;

}

}

t++;

}

return 0;

}