UVa 11732-strcmp () anyone?

Title: Give you some words, give you a string comparison function, all words are parity, what is the number of comparisons.

Analysis: Strings, Terry.

First of all. Look at the data size, assuming normal discovery clues, meeting tle and MLE.

As a result, the normal dictionary tree depth is 1000, and there may be a lot of space wasted.

So, using a tree compression algorithm (left brother, right child). can improve execution efficiency.

And then. You can do it in the dictionary tree. Statistics, you can make a contribution to the statistics, or the edge of the achievement of statistics.

Here I choose the edge of the method of statistical methods (online a large number of practices, are the first achievements and then statistics, search solution)

Each time you insert a word. It is only compared to the word inserted earlier, and the number of words is the same as the number of each letter.

The number of times each letter in a word is compared. Is the number of words that the root node of the letter contains.

The word comparison functions such as the following:

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

Due to the comparison function, the following points should be noted when calculating:

1. Two words of the same length of L are more expensive than 2l-1. The final extrapolation of the S-end;

2. The comparison length of the word should be strlen (str) +1, The Terminator is also a relatively more ring;

3. Assume that two words are the same, then compute an end inference more than once. +1 more times.

Note: Use ll, otherwise the data will overflow.

#include <iostream> #include <cstdlib> #include <cstring> #include <cstdio>using namespace std;    typedef long LONG Ll;char words[1010];/* Trie define */#define NODESIZE 4444444//number of nodes typedef struct NODE1 {Char    Value;int size;int Count; Node1* lchild;node1* Rchild;}  Tnode;  Tnode Dict[nodesize];        Class Trie {private:ll count;          int size;    tnode* Root;           Public:trie () {initial ();}              void Initial () {memset (dict, 0, sizeof (dict));        Size=0;count=0ll;root=newnode (0);          } tnode* NewNode (char c) {Dict[size].value = C;return &dict[size + +];}             void Insert (char* word, int L) {tnode* now = Root->rchild,*save = root;            int same = 1;  for (int i = 0; I <= L; + + i) {if (!now) {save->rchild = NewNode (Word[i]); now = Save->rchild;now->count = 1;same = 0;} else {if (i) count + = Now->count;count + = nOw->count ++;while (now->lchild && now->value! = word[i]) now = now->lchild;if (now->value! = Wor D[i]) {now->lchild = NewNode (Word[i]); now = Now->lchild;same = 0;}}            Save = Now;now = save->rchild;            } if (same) save->size + +;        Count + = save->size; } LL query () {return count;}}  Trie /* Trie END */int main () {int case = 1,n;while (scanf ("%d", &n)! = EOF) {if (! N) break;trie.initial (); for (int i = 0; i < N; + + i) {scanf ("%s", words); Trie.insert (Words,strlen (words));} printf ("Case%d:%lld\n", Case ++,trie.query ());} return 0;}

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UVa 11732-strcmp () anyone?