UVA 10029 Edit Step Ladders (DP)

Problem C:edit Step Ladders

An edit "is a transformation" from "one word X to another word y such" x and y are words in the dictionary, and x CA n is transformed to Y by adding, deleting, or changing one letter. So the transformation from dig to dog or from dog to did are edit both. An edit step Ladder is a lexicographically ordered sequence of words W1, w2, ... wn such that the transformation from WI t o wi+1 is a edit step for all I from 1 to n-1.

For a given dictionary, your are to compute the length of the longest edit step ladder.

Input

The input to your program consists of the dictionary-a set of lower case words in lexicographic order-one per line. No word exceeds letters and there are no more than 25000 in the words.

Output

The output consists of a single integer, the number of words in the longest edit step ladder.

Sample Input

Cat
dig
dog
Fig
fin
fine
fog
Log
wine

Sample Output

5

Given a dictionary, if deleted, add or replace can be converted into another word, then these two words for the ladder change, find the longest step changes in the dictionary.

Idea: DP. Lis, with the n^2 timeout ... And then don't want to understand the solution, found that there are N (Logn) methods.

All the words can be transformed out of the case, and then in front of the 2-point search, because the sequence itself is a dictionary order is feasible. So we can get through this.

Code:

 #include <stdio.h> #include <string.h> int max (int a, int b) {return a > B. a:b} int min (int a, I
NT B) {return a < b a:b;} const int N = 25555, M = 20;
Char Word[n][m], str[m];
  
int n, I, J, K, L, dp[25555], ans;
    void Change (char *word, Char *str, char c, int wei) {int i;
    for (i = 0; word[i]; i + +) str[i] = Word[i];
    Str[wei] = c;
  
Str[i] = ' the ';
    } void del (char *word, char *str, int wei) {int i;
    for (i = 0; I < Wei i + +) str[i] = Word[i];
    for (i = wei + 1; word[i]; i + +) str[i-1] = Word[i];
Str[i-1] = ' the ';
    } void Add (char *word, Char *str, char c, int wei) {int i;
    for (i = 0; I < Wei i + +) str[i] = Word[i];
    Str[wei] = c;
    for (i = Wei; word[i]; i + +) Str[i + 1] = Word[i];
Str[i + 1] = ';
    } void Tra (char *word, Char *str, char c, int wei, int flag) {if (flag = 0) Change (Word, str, C, Wei); else if (flag = = 1) del (Word, str,Wei);
else Add (Word, str, C, Wei);
    int find (char *str, int j) {int i = 0;
    J--;
        while (I <= j) {int mid = (i + j)/2;
        if (strcmp (Word[mid], str) = = 0) {return mid;
        else if (strcmp (Word[mid], str) < 0) {i = mid + 1;
    else J = mid-1;
} return-1;
    int main () {while (gets (Word[n])) {n + +;
    Ans = 0;
        for (i = 0; i < n; i + +) {Dp[i] = 1;
                    for (k = 0; k < 3; k + +) {for (j = 0; word[i][j]; j + +) {for (L = 0; L < L + +) {
                    TRA (word[i], str, ' a ' + L, J, K);
                    if (strcmp (word[i], str) < 0) break;
                    int mid = find (str, i);
                If (mid >= 0) dp[i] = max (Dp[i], Dp[mid] + 1);
    ans = max (ans, dp[i]);
    printf ("%d\n", ans);
return 0; }

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