strcmp() Anyone? Time Limit:2000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

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UVA 11732 Appoint description:

Description

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:

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.

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

字符串很多，数据量很大，两两比较肯定超时，如果把所有的单词加入到一棵Trie树，然后比较就只需要比较 树出现分叉的地方并且获得那个地方节点的权值，比如 than 和that比较，首先加入Trie树，如图所示

than和that的前三个字符相同，第四个不同，因此比较次数为7，即n（n-1）/2次

// UVa11732 strcmp() Anyone?
// Rujia Liu
#include<cstring>
#include<vector>
#include<cstdio>
using namespace std;
#define LL long long
const int maxnode = 4000 * 1000 + 10;
const int sigma_size = 26;
int head[maxnode]; /// head[i]为第i个结点的左儿子编号
int next[maxnode]; /// next[i]为第i个结点的右兄弟编号
char ch[maxnode];  /// ch[i]为第i个结点上的字符
int tot[maxnode];  /// tot[i]为第i个结点为根的子树包含的叶结点总数
int flag[maxnode];
int sz;/// 结点总数
LL ans;
void insert(const char *s)
{
    ans+=tot[0]; /// 答案
    tot[0]++;
    //sz = 1;    /// 初始时只有一个根结点

    int u = 0, v, n = strlen(s);
    int k;
/// 插入字符串s（包括最后的'\0'），沿途更新tot

    for(int i = 0; i < n; i++)
    {
        ///找字符a[i]
        bool found = false;
        for(v = head[u]; v != 0; v = next[v])
            if(ch[v] == s[i])   /// 找到了
            {
                found = true;
                break;
            }
        if(!found)
        {
            v = sz++; /// 新建结点
            tot[v] = 0;
            flag[v]=0;
            ch[v] = s[i];
            next[v] = head[u];
            head[u] = v; /// 插入到链表的首部
            head[v] = 0;
        }
        u = v;
        ans+=tot[u]*2;
        tot[u]++;
    }
    k=v;
/// 统计LCP=u的所有单词两两的比较次数之和

    if(flag[k])
        ans+=flag[k];
    flag[k]++;
}


const int maxl = 1000 + 10;/// 每个单词最大长度
int n;
char word[maxl];
int main()
{
    int kase = 1;
    while(scanf("%d", &n) == 1 && n)
    {
        //trie.clear();
        ans=0,sz=1;
        tot[0] = head[0] = next[0] = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%s", word);
            insert(word);
        }
        printf("Case %d: %lld\n", kase++, ans);
    }
    return 0;
}