Uva 11732 strcmp() Anyone?

 

 

strcmp() Anyone?

Time Limit: 2000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

J	「strcmp()」 Anyone? Input: Standard Input Output: Standard Output

 

strcmp() is a library function in C/C++ which compares two strings. It takes two strings as input parameter and decides which one is lexicographically larger or smaller: If the first string is greater then it returns a positive value, if the second string is greater it returns a negative value and if two strings are equal it returns a zero. The code that is used to compare two strings in C/C++ library is shown below:

int strcmp(char *s, char *t) {     int i;     for (i=0; s[i]==t[i]; i++)         if (s[i]=='\0')             return 0;     return s[i] - t[i]; }

Figure: The standard strcmp() code provided for this problem.

 

The number of comparisons required to compare two strings in strcmp() function is never returned by the function. But for this problem you will have to do just that at a larger scale. strcmp() function continues to compare characters in the same position of the two strings until two different characters are found or both strings come to an end. Of course it assumes that last character of a string is a null (‘\0’) character. For example the table below shows what happens when 「than」 and 「that」; 「therE」 and 「the」 are compared using strcmp() function. To understand how 7 comparisons are needed in both cases please consult the code block given above.

 

t

h

a

N

\0

t

h

e

r

E

\0

=

=

=

≠

=

=

=

≠

t

h

a

T

\0

t

h

e

\0

Returns negative value 7 Comparisons

Returns positive value 7 Comparisons

 

Input

The input file contains maximum 10 sets of inputs. The description of each set is given below:

 

Each set starts with an integer N (0<N<4001) which denotes the total number of strings. Each of the next N lines contains one string. Strings contain only alphanumerals (‘0’… ‘9’, ‘A’… ‘Z’, ‘a’… ‘z’) have a maximum length of 1000, and a minimum length of 1.  

 

Input is terminated by a line containing a single zero. Input file size is around 23 MB.

 

Output

For each set of input produce one line of output. This line contains the serial of output followed by an integer T. This T denotes the total number of comparisons that are required in the strcmp() function if all the strings are compared with one another exactly once. So for N strings the function strcmp() will be called exactly  times. You have to calculate total number of comparisons inside the strcmp() function in those  calls. You can assume that the value of T will fit safely in a 64-bit signed integer. Please note that the most straightforward solution (Worst Case Complexity O(N² *1000)) will time out for this problem.

 

Sample Input                              Output for Sample Input

2 a b 4 cat hat mat sir 0

Case 1: 1 Case 2: 6

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Problem Setter: Shahriar Manzoor, Special Thanks: Md. Arifuzzaman Arif, Sohel Hafiz, Manzurur Rahman Khan

 

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Trie，由於結點種類太多，因此要改用左兒子右兄弟的Trie，否則會TLE。

判斷比較幾回的時候，要考慮字符串徹底相同的狀況。

1：規律是 sum( node *(node-1))   +  n*(n-1)/2  + sum ( flag*(flag-1)/2 )

即  區分開的次數+每一個通過的結點比較的次+重複的串在結尾比較了2次

2：也能夠每一個節點分開來考慮，累計單詞在每一個節點分開的時候，所比較的次數（第一個節點分開的單詞要比較1次，第二個節點分開的要比較3次。。。。）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 const int maxnode=4002*1002;
 8 
 9 typedef long long int LL;
10 
11 struct Trie
12 {
13     int left[maxnode],right[maxnode],val[maxnode],cnt;
14     char ch[maxnode];
15     LL ans;
16     Trie(){}
17     void init()
18     {
19         cnt=1;
20         left[0]=right[0]=val[0]=ch[0]=ans=0;
21     }
22     void insert(const char * str)
23     {
24         int len=strlen(str);
25         int u=0,j;
26         for(int i=0;i<=len;i++)
27         {
28             for(j=left[u];j;j=right[j])
29             {
30                 if(ch[j]==str[i]) break;
31             }
32             if(j==0)
33             {
34                 j=cnt++;
35                 right[j]=left[u];
36                 left[u]=j;
37                 left[j]=0;ch[j]=str[i];
38                 val[j]=0;
39             }
40            // printf("%d:%c  %d * %d =  %d\n",i,str[i],val[u]-val[j],(i<<1|1),(val[u]-val[j])*(i<<1|1));
41             ans+=(val[u]-val[j])*(i<<1|1);
42             if(i==len)
43             {
44            //     printf("%d:%c  %d * %d =  %d\n",i,str[i],val[j],(i+1)<<1,val[j]*((i+1)<<1));
45                 ans+=val[j]*((i+1)<<1);
46                 val[j]++;
47             }
48             val[u]++;u=j;
49         }
50     }
51 }tree;
52 
53 int main()
54 {
55     char dic[2000];
56     int n,cas=1;
57     while(scanf("%d",&n)!=EOF&&n)
58     {
59         tree.init();
60         while(n--)
61         {
62             scanf("%s",dic);
63             tree.insert(dic);
64         }
65         printf("Case %d: %lld\n",cas++,tree.ans);
66     }
67     return 0;
68 }