644 - Immediate Decodability（AC）

该题有ANSI C写了一遍，通过使用二叉树的思想可以很好的解决这个问题，该题是用UVa系统测评的，发现如果用标准的C写代码的话，语法要求真的

很严格，提交了几次都是语法错误，单行注释都报错，最要紧的是连内建的bool类型都没有，所以我用int型代替，下面是AC的代码

#include <stdio.h>
#define Nil -1
char left[50], right[50], tree_index[50];
char code[40];
int top = 0;
int tree_root = -1;
char* out_buff_is = "Set %d is immediately decodable\n";
char* out_buff_not = "Set %d is not immediately decodable\n";

int inset_tree(char* buff)
{
	char *p = buff;
	int parent;
	int child = tree_root;
	int is_new_node = 1;
	for( ; *p ; ++p)
	{
		parent = child;
		if(!is_new_node && left[child] == Nil && right[child] == Nil && child != tree_root) return 0;
		switch(*p)
		{
		case '0':
			if(left[child] == Nil)
			{
				child = top++;
				left[parent] = child;
				right[child] = Nil;
				left[child] = Nil;
				code[child] = *p;
				is_new_node = 1;
			}
			else
			{
				child = left[child];
				is_new_node = 0;
			}
			break;
		case '1':
			if(right[child] == Nil)
			{
				child = top++;
				right[parent] = child;
				right[child] = Nil;
				left[child] = Nil;
				code[child] = *p;
				is_new_node = 1;
			}
			else
			{
				child = right[child];
				is_new_node = 0;
			}
			break;
		}
	}
	return is_new_node;
}

int main()
{
	char buff[20];
	int is_decodable;
	int num_case;
	is_decodable = 1;
	num_case = 1;
	tree_root = top++;
	left[tree_root] = Nil;
	right[tree_root] = Nil;
	code[tree_root] = '-';
	while(gets(buff)){
		if(buff[0] == '9')
		{
			if(is_decodable) printf(out_buff_is,num_case);
			else			 printf(out_buff_not,num_case);
			++num_case;
			top = 1;
			left[tree_root] = Nil;
			right[tree_root] = Nil;
			is_decodable = 1;
		}
		else if(is_decodable == 0) continue;
		if(!inset_tree(buff))
		{
			is_decodable = 0;
		}
	}
	return 0;
}