Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 14371 | Accepted: 6862 |
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)
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 standard input. 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.
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
Source
Regionals 1998 >> North America - Pacific Northwest
问题链接:POJ1056 HDU1305 ZOJ1808 UVA644 UVALive5479 Immediate Decodability
问题描述:
输入若干0和1组成的字符串集合,字符串“9”为字符串输入结束,判断其中的某个字符串是否为另外一个字符串的前缀。
问题分析:
这个问题原理上来说,看似一个Huffman编码问题,也可以说是。
该问题用构建字典树的方法来解决。构造一个二叉字典树进行判断即可。
程序说明:(略)
参考链接:(略)
题记:(略)
AC的C++语言程序如下:
/* POJ1056 HDU1305 ZOJ1808 UVA644 UVALive5479 Immediate Decodability */
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 8;
const int LEN = 10;
const int SIZE = 2;
const char SCHAR = '0';
struct Tree {
int acnt; // access count
int ccnt; // child number
int childs[SIZE];
void init()
{
acnt = 1;
ccnt = 0;
memset(childs, 0, sizeof(childs));
}
} trie[N * LEN];
int ncnt; // Trie Node count
char s[LEN]; // Input
bool insert(char s[])
{
int p = 0;
for(int i = 0; s[i]; i++) {
int k = s[i] - SCHAR;
int child = trie[p].childs[k];
if(child) {
trie[child].acnt++;
p = child;
} else {
if(i > 0 && trie[p].acnt > 1 && trie[p].ccnt == 0)
return false;
trie[++ncnt].init();
trie[p].childs[k] = ncnt;
trie[p].ccnt++;
p = ncnt;
}
}
return trie[p].ccnt == 0;
}
int query(char s[])
{
int p = 0;
for (int i = 0; s[i]; i++) {
int k = s[i] - SCHAR;
int child = trie[p].childs[k];
if (child == 0)
return 0;
else
p = child;
}
return trie[p].ccnt;
}
int main()
{
int caseno = 0;
bool flag = true;
ncnt = 0;
trie[ncnt].init();
while(~scanf("%s", s)) {
if(s[0] == '9') {
if(flag)
printf("Set %d is immediately decodable\n", ++caseno);
else
printf("Set %d is not immediately decodable\n", ++caseno);
flag = true;
ncnt = 0;
trie[ncnt].init();
} else {
if(flag)
if(!insert(s))
flag = false;
}
}
return 0;
}