HDOJ 1671 Phone List(ディクショナリツリー+リリーススペース)
3471 ワード
Phone List
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15218 Accepted Submission(s): 5136
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
Sample Output
問題:あなたにn個の電話番号をあげて、1つの索引によって1人の構想を探し当てることができますか:実はn個の列の間に子列関係があるかどうかを調べて、最も長い列を第1個の列にして、それから後ろの列でそれと比較してまた、最後に空間を釈放することを忘れてはいけなくて、さもなくばメモリを超えることができます
acコード:
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15218 Accepted Submission(s): 5136
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
問題:あなたにn個の電話番号をあげて、1つの索引によって1人の構想を探し当てることができますか:実はn個の列の間に子列関係があるかどうかを調べて、最も長い列を第1個の列にして、それから後ろの列でそれと比較してまた、最後に空間を釈放することを忘れてはいけなくて、さもなくばメモリを超えることができます
acコード:
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
#include<iostream>
#include<algorithm>
#define fab(a) (a)>0?(a):(-a)
#define LL long long
#define MAXN 10010
#define mem(x) memset(x,0,sizeof(x))
#define INF 0xfffffff
using namespace std;
struct s
{
int num;
s *next[10];
};
struct word
{
char phone[12];
int len;
}a[MAXN];
s *root;
bool cmp(word q,word w)
{
return q.len<w.len;
}
void create(char *str)
{
int len=strlen(str);
s *p=root,*q;
for(int i=0;i<len;i++)
{
int id=str[i]-'0';
if(p->next[id]==NULL)
{
q=(s *)malloc(sizeof(s));
q->num=1;
for(int j=0;j<10;j++)
q->next[j]=NULL;
p->next[id]=q;
p=p->next[id];
}
else
{
p->next[id]->num++;
p=p->next[id];
}
}
//p->num=-1;
}
int find(char *ss)
{
int len=strlen(ss);
s *p=root;
for(int i=0;i<len;i++)
{
int id=ss[i]-'0';
p=p->next[id];
if(p==NULL)
return 0;
// if(p->num==-1)
// return -1;
}
return p->num;
}
void begin()
{
for(int i=0;i<10;i++)
root->next[i]=NULL;
}
void freetree(s *t)
{
if(t==NULL)
return;
for(int i=0;i<10;i++)
{
if(t->next[i]!=NULL)
freetree(t->next[i]);
}
free(t);
return;
}
int main()
{
int t,i,n;
scanf("%d",&t);
while(t--)
{
int bz=0;
root=(s *)malloc(sizeof(s));
begin();
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",a[i].phone);
int l=strlen(a[i].phone);
a[i].len=l;
}
sort(a,a+n,cmp);
create(a[n-1].phone);
for(i=n-2;i>=0;i--)
{
if(find(a[i].phone)==0)
create(a[i].phone);
else
{
bz=1;;
break;
}
}
printf(bz?"NO
":"YES
");
freetree(root);
}
return 0;
}