UVA - 10029 Edit Step Ladders （二分+hash）

Description

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, ... wn such that the transformation from
wi to wi+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：给你一个递增的字符串数组，给你三种操作方法变成其它的串，问你最长的可能
思路：hash+二分，dp[i]表示从i開始的串的最长变化可能。记忆化搜索
[code]#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 25010;
const int HASH = 1000010;

int n, head[HASH], next[MAXN], f[MAXN];
char b[MAXN][20], temp[20];

int hash(char *s) {
int v = 0,seed = 131;
while (*s)
v = v * seed + *(s++);
return (v & 0x7fffffff) % HASH;
}

void insert(int s) {
int h = hash(b[s]);
next[s] = head[h];
head[h] = s;
}

int search(char *s) {
int i,h = hash(s);
for (i = head[h]; i != -1; i = next[i])
if (!strcmp(b[i],s))
break;
return i;
}

void add(char *s, int p, int d) {
int i = 0, j = 0;
while (i < p)
temp[j++] = s[i++];
temp[j++] = 'a' + d;
while (s[i])
temp[j++] = s[i++];
temp[j] = '\0';
}

void del(char *s, int p) {
int i = 0,j = 0;
while (i < p)
temp[j++] = s[i++];
i++;
while (s[i])
temp[j++] = s[i++];
temp[j] = '\0';
}

void change(char *s, int p, int d) {
strcpy(temp, s);
temp[p] = 'a' + d;
}

int dp(int s) {
if (f[s] != -1)
return f[s];
int ans = 1;
int len = strlen(b[s]);
for (int p = 0; p <= len; p++)
for (int d = 0; d < 26; d++) {
add(b[s], p, d);
int v = search(temp);
if (v != -1 && strcmp(b[s], temp) < 0){
int t = dp(v);
if (ans < t+1)
ans = t+1;
}
}
for (int p = 0; p < len; p++) {
del(b[s], p);
int v = search(temp);
if (v != -1 && strcmp(b[s], temp) < 0) {
int t = dp(v);
if (ans < t+1)
ans = t+1;
}
}
for (int p = 0; p < len; p++)
for (int d = 0; d < 26; d++) {
change(b[s], p, d);
int v = search(temp);
if (v != -1 && strcmp(b[s], temp) < 0) {
int t = dp(v);
if (ans < t+1)
ans = t+1;
}
}
return f[s] = ans;
}

int main() {
n = 0;
memset(head, -1, sizeof(head));
while (scanf("%s", b
) != EOF) {
insert(n),++n;
}
memset(f, -1, sizeof(f));
int ans = 0;
for (int i = 0; i < n; i++) {
int t = dp(i);
if (ans < t)
ans = t;
}
printf("%d\n", ans);
return 0;
}

[/code]