UVA 11732 —— 左儿子右兄弟表示法&&Tire

11732 - strcmp() Anyone?

Time limit: 2.000 seconds

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

Case 1: 1 Case 2: 6

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

参考了一下lrj的代码，感觉左儿子右兄弟表示法相当好用，赞！

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>

using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1

#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 4000 * 1000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-4
const int MOD = (int)1e9 + 7;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
struct Trie
{
    int head[MAXN];/// head[i]为第i个结点的左儿子编号
    int next[MAXN];/// next[i]为第i个结点的右兄弟编号
    int tot[MAXN];/// ch[i]为第i个结点上的字符
    char ch[MAXN]; /// tot[i]为第i个结点为根的子树包含的叶结点总数
    int sz; /// 结点总数
    LL ans;
    void clear()
    {
        sz = 1 , head[0] = next[0] = tot[0] = 0;
    }
    /// 插入字符串s（包括最后的'\0'），沿途更新tot
    void insert(char * s)
    {
        int u = 0 , v , n = strlen(s);
        tot[0] ++;
        FORR(i ,0 , n)
        {
            bool found = false;
            for(v = head[u] ; v != 0 ; v = next[v])
            {
                if(ch[v] == s[i])
                {
                    found = true ;
                    break;
                }
            }
            if(!found)
            {
                v = sz++;/// 新建结点
                tot[v] = 0;
                ch[v] = s[i];
                next[v] = head[u];
                head[u] = v;
                head[v] = 0;
            }
            u = v;
            tot[u] ++;
        }

    }
    /// 统计LCP=u的所有单词两两的比较次数之和
    void dfs(int depth , int u)
    {
        if(head[u] == 0)
        {
            ans += tot[u] * (tot[u] - 1) * depth;
        }
        else
        {
            int sum = 0;
            for(int v = head[u] ; v != 0 ; v = next[v])
            {
                sum += tot[v] * (tot[u] - tot[v]);/// 子树v中选一个串，其他子树中再选一个
            }
            ans += sum / 2 * (2 * depth + 1);/// 除以2是每种选法统计了两次
            for(int v = head[u] ; v != 0 ; v = next[v])dfs(depth + 1 , v);
        }
    }
    LL count()
    {
        ans =0 ;
        dfs(0 , 0);
        return ans;
    }
};

char str[1010];
Trie trie;
int n;
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    //freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
#endif // Online_Judge
    int kase = 1;
    while(~scanf("%d" , &n) , n)
    {
        trie.clear();
        while(n--)
        {
            scanf("%s" ,str);
            trie.insert(str);
        }

        printf("Case %d: %lld\n" , kase++ , trie.count());
    }
    return 0;
}