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，因为结点种类太多，所以要改用左儿子右兄弟的Trie，不然会TLE。

判断比较几次的时候，要考虑字符串完全相同的情况。

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

即  区分开的次数+每个经过的结点比较的次+重复的串在结尾比较了2次

2：也可以每个节点分开来考虑，累计单词在每个节点分开的时候，所比较的次数（第一个节点分开的单词要比较1次，第二个节点分开的要比较3次。。。。）

 #include <iostream>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 const int maxnode=*;

 typedef long long int LL;

 struct Trie
 {
     int left[maxnode],right[maxnode],val[maxnode],cnt;
     char ch[maxnode];
     LL ans;
     Trie(){}
     void init()
     {
         cnt=;
         left[]=right[]=val[]=ch[]=ans=;
     }
     void insert(const char * str)
     {
         int len=strlen(str);
         int u=,j;
         for(int i=;i<=len;i++)
         {
             for(j=left[u];j;j=right[j])
             {
                 if(ch[j]==str[i]) break;
             }
             if(j==)
             {
                 j=cnt++;
                 right[j]=left[u];
                 left[u]=j;
                 left[j]=;ch[j]=str[i];
                 val[j]=;
             }
            // printf("%d:%c  %d * %d =  %d\n",i,str[i],val[u]-val[j],(i<<1|1),(val[u]-val[j])*(i<<1|1));
             ans+=(val[u]-val[j])*(i<<|);
             if(i==len)
             {
            //     printf("%d:%c  %d * %d =  %d\n",i,str[i],val[j],(i+1)<<1,val[j]*((i+1)<<1));
                 ans+=val[j]*((i+)<<);
                 val[j]++;
             }
             val[u]++;u=j;
         }
     }
 }tree;

 int main()
 {
     char dic[];
     int n,cas=;
     while(scanf("%d",&n)!=EOF&&n)
     {
         tree.init();
         while(n--)
         {
             scanf("%s",dic);
             tree.insert(dic);
         }
         printf("Case %d: %lld\n",cas++,tree.ans);
     }
     return ;
 }