UVA 10029 Edit Step Ladders(dp)

Problem C: Edit Step Ladders

An  edit step  is a transformation from one word  x  to another word  y  such that  x  and  y  are words in the dictionary, and  x  can be transformed to  y  by adding, deleting, or changing one letter. So the transformation from  dig  to  dog  or from  dog  to  do  are both edit steps. An  edit step ladder  is a lexicographically ordered sequence of words  w₁, w₂, ... w_n  such that the transformation from  w_i  to  w_i+1  is an edit step for all  i  from 1 to  n-1 .

For a given dictionary, you 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 16 letters and there are no more than 25000 words in the dictionary.

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

题意：给定一个字典，如果经过删除，添加或替换能转换成另一个单词，则这两个单词为阶梯变化，求出最长阶梯变化的字典。

思路：dp。lis，用n^2 超时了。。。然后没想明白看题解，发现有n(logn)方法。

把每个单词可以变换的情况全部弄出来，然后在前面进行2分查找，由于序列本身是字典序所以可行。这样就能过了

代码：

#include <stdio.h>
#include <string.h>
int max(int a, int b) {return a > b ? a : b;}
int min(int a, int 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] = '\0';

}

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] = '\0';
}

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] = '\0';
}

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 < 26; 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;
}