UVA11732 strcmp() Anyone? Trie

最新推荐文章于 2023-03-24 10:52:21 发布

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本文介绍了一种使用字典树优化字符串比较的方法，该方法能够有效减少strcmp()函数在大规模字符串比较时所需的比较次数。通过构建字典树并在建树过程中进行计算，可以显著提高字符串比较的效率。

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简单的字典树应用，建树的时候就可以进行计算了。注意字符串结束符也会比较。不要漏掉。当串完全相同的时候最好自己出几组数据，这个地方是这道题的难点。其他还好。注意有大小写还有阿拉伯数字。

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.

≠

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

cat

hat

mat

sir

Case 1: 1

Case 2: 6

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

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cstdlib>

using namespace std;

#define MAXN 4010*1000
#define MAX 65

int sz,t[MAXN][MAX];
int jud[MAXN];
int fin[MAXN];

void clear(){
    sz=1;
    memset(t[0],-1,sizeof(t[0]));
    jud[0]=0;
    fin[0]=0;
}

int idx(char c){
    if(c<='9'&&c>='0')
        return c-'0';
    if(c<='Z'&&c>='A')
        return c-'A'+11;
    return c-'a'+37;
}

long long ret;

void insert(char *s){
    int u=0;
    int n=strlen(s);
    ret+=jud[0];
    jud[0]++;
    for(int i=0;i<n;i++){
        int c=idx(s[i]);
       // cout<<t[u][c]<<" "<<u<<" "<<c<<endl;
        if(t[u][c]==-1){
            memset(t[sz],-1,sizeof(t[sz]));
            jud[sz]=0;
            fin[sz]=0;
            t[u][c]=sz++;
        }
        u=t[u][c];
        ret+=jud[u];
        ret+=jud[u];
        jud[u]++;
    }
    ret+=fin[u];
    fin[u]++;
}

int n;
char str[2000];

int main(){
    int cs=1;
    while(~scanf("%d",&n)&&n){
        ret=0;
        clear();
        for(int i=0;i<n;i++){
            scanf("%s",str);
            insert(str);
        }
        printf("Case %d: %lld\n",cs++,ret);
    }
    return 0;
}