sdutoj 2151 Phone Number

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2151

 

Phone Number

 

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

We know that if a phone number A is another phone number B’s prefix, B is not able to be called. For an example, A is 123 while B is 12345, after pressing 123, we call A, and not able to call B.
Given N phone numbers, your task is to find whether there exits two numbers A and B that A is B’s prefix.
 

输入

  The input consists of several test cases.
 The first line of input in each test case contains one integer N (0<N<1001), represent the number of phone numbers.
 The next line contains N integers, describing the phone numbers.
 The last case is followed by a line containing one zero.

输出

  For each test case, if there exits a phone number that cannot be called, print “NO”, otherwise print “YES” instead.

示例输入

2

012

012345

2

12

012345

0

示例输出

NO

YES

提示

 

来源

 2010年山东省第一届ACM大学生程序设计竞赛

示例程序

 

 

AC代码：

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cstdlib>

 5 using namespace std;

 6 

 7 struct node

 8 {

 9     int count;

10     node *child[10];

11     node()

12     {

13         count=0;

14         int i;

15         for(i=0; i<10; i++)

16             child[i]=0;

17     }

18 };

19 

20 node *current, *newnode;

21 

22 void insert(node *root, char *ss)

23 {

24     //printf("%s*****\n",ss);

25     int i, m;

26     current=root;

27     for(i=0; i<strlen(ss); i++)

28     {

29         m=ss[i]-'0';

30         if(current->child[m]!=NULL)

31         {

32             current=current->child[m];

33             ++(current->count);

34         }

35         else

36         {

37             newnode=new node;

38             ++(newnode->count);

39             current->child[m]=newnode;

40             current=newnode;

41         }

42     }

43 }

44 

45 int search(node *root, char *ss)

46 {

47     //printf("%s*****\n",ss);

48     int i, m;

49     current=root;

50     for(i=0; i<strlen(ss); i++)

51     {

52         m=ss[i]-'0';

53         if(current->child[m]==NULL)

54             return 0;

55         current=current->child[m];

56     }

57     return current->count;

58 }

59 

60 int main()

61 {

62     char str[30], s[1010][30];

63     int t, flag, i;

64     while(scanf("%d",&t))

65     {

66         if(t==0) break;

67         flag=0;

68         node *root=new node;

69         for(i=0; i<t; i++)

70         {

71             scanf("%s",str);

72             strcpy(s[i], str);

73             //puts(s[i]);

74             insert(root, str);

75         }

76         for(i=0; i<t; i++)

77         {

78             if(search(root, s[i])-1)

79             {

80                 flag=1;

81                 break;

82             }

83         }

84         if(flag==0) printf("YES\n");

85         else printf("NO\n");

86     }

87     return 0;

88 }

View Code