hdoj 1671字典树水题之三 静态数组节约内存法

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6265    Accepted Submission(s): 2131

Problem Description

Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let’s say the phone catalogue listed these numbers:
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it’s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob’s phone number. So this list would not be consistent.

 

Input

The first line of input gives a single integer, 1 <= t <= 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

 

Output

For each test case, output “YES” if the list is consistent, or “NO” otherwise.

 

Sample Input

2

3

911

97625999

91125426

5

113

12340

123440

12345

98346

 

Sample Output

NO

YES

 

题目思路：主要判断是否有相互的前缀，加一个标记即可。

注意：如果单纯的用malloc去申请内存，内存会超出，这里用一个数组去存，每个集体，从0开始，这样就节约了内存。如果用free或者move会浪费时间！

代码：

#include<stdio.h>
#include<malloc.h>
#include<string.h>
typedef struct node
{
	int end;
	struct node *next[10];
}node;
int k;
node m[1000001];//

int insert(char *str,node *T)
{
	node *p,*q;
	int len,flag=0,id,i,j;
	p=T;
	len=strlen(str);
	for(i=0;i<len;++i)
	{
		id=str[i]-'0';
		if(p->end==1)
			return 1;//判断是否存在较短的串是他的前缀
		if(p->next[id]==NULL)
		{
		//	q=(node*)malloc(sizeof(node));
		//	node *q=new node();
			q=&m[k++];
			for(j=0;j<10;j++)
			{
				q->end=0;
				q->next[j]=NULL;
			}
			p->next[id]=q;
		}
	/*	if(i==len-1)
			p->end=1;*/
		p=p->next[id];
	}
	for(i=0;i<10;++i)
		if(p->next[i]!=NULL)
			return 1;//判断是否有比他长的串
	p->end=1;
	return flag;
}


int main()
{
	int t,n,flag,i;
	char str[11];
	node *T;
	scanf("%d",&t);
	while(t--)
	{
		//node *T=new node();
		k=0;
		T=&m[k++];
		//T=(node*)malloc(sizeof(node));
		for(i=0;i<10;++i)
		{
			T->end=0;
			T->next[i]=NULL;
		}
		scanf("%d",&n);
		flag=0;
		while(n--)
		{
			scanf("%s",&str);
			if(insert(str,T))
				flag=1;
		}
		if(flag)
			printf("NO\n");
		else
			printf("YES\n");
		//free(T);
	}
	return 0;
}