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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 wordy 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 fromdig to dog or from dog to do are both edit steps. Anedit step ladder is a lexicographically ordered sequence of words w1, w2, ... wn such that the transformation fromwi to wi+1 is an edit step for all i from 1 ton-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他的串,问你最长的可能思路:hash+二分,dp[i]表示从i开始的串的最长变化可能,记忆化搜索#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[n]) != 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; }
UVA - 10029 Edit Step Ladders (二分+hash)
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