POJ1056 IMMEDIATE DECODABILITY【数据结构】

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)

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).

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.

01
10
0010
0000
9
01
10
010
0000
9

Set 1 is immediately decodable
Set 2 is not immediately decodable
 

Pacific Northwest 1998

将各个编码序列作为二叉树的节点序列建立二叉树，并在过程中标记编码序列的最后一位，然后遍历二叉树，如果存在非叶节点的istail为1，即可说明存在某序列为另一序列的前缀序列。

#include <stdio.h>
#include <stdlib.h>

typedef struct btree{
	int istail;
	struct btree * left;
	struct btree * right;
}BTree, *pBTree;

char data[11];
BTree * root = NULL;

void insert(char data[]){
	int i= 0;
	BTree * p = NULL;
	if (root == NULL){
		root = (BTree *)malloc(sizeof(BTree));
		root->istail = 0;
		root->left = root->right = NULL;
	}
	p = root;
	while (data[i] != '\0'){
		if (data[i] == '0'){
			if (p->left != NULL){
				p = p->left;
			} else{
				p->left = (BTree *)malloc(sizeof(BTree));
				p = p->left;
				p->istail = 0;
				p->left = p->right = NULL;
			}
		} else{
			if (p->right != NULL){
				p = p->right;
			} else {
				p->right = (BTree *)malloc(sizeof(BTree));
				p = p->right;
				p->istail = 0;
				p->left = p->right = NULL;
			}
		}
		++i;
	}
	p->istail = 1;
}

int isImmediately(BTree * root){
	BTree * p = root;
	while (p != NULL){
		if (p->istail == 1 && (p->left != NULL || p->right != NULL))
			return 0;
		else
			return isImmediately(p->left) && isImmediately(p->right);
	}
	return 1;
}

void destoryBTree(pBTree * root){
		if ((*root)->left)
			destoryBTree(&(*root)->left);
		if ((*root)->right)
			destoryBTree(&(*root)->right);
		free(*root);
		*root = NULL;
}

int main(void){
	int count = 0;
	while (gets(data)){
		if (data[0] == '9'){	
			if (isImmediately(root))
				printf("Set %d is immediately decodable\n", ++count);
			else
				printf("Set %d is not immediately decodable\n", ++count);
			destoryBTree(&root);
		} else{
			insert(data);
		}
	}

	return 0;
}