POJ 3267-The Cow Lexicon 动态规划

题目来源: http://acm.pku.edu.cn/JudgeOnline/problem?id=3267

 

解题报告:

 

一道动态规划的题，一开始的思路有些错，看了网上的解题报告后，才理正了思路，看来还是要多做题，纸上得来终觉浅！

 

这道题求的是某个字符串recmsg，至少删去多少个字符后，剩下的字符串是由dict中的单词可以组成的。

 

对字符串recmsg[0...L-1]，设dp[i]＝m，代表recmsg[i...L-1]中至少删去m个字符后，剩下的字符串是由dict中的单词可以组成的。则问题求的就是dp[0]。且可以得到递归式：

 

dp[i]=dp[i+1]+1，如果dict中不存在单词可以与recmsg[i...L-1]中以recmsg[i]开头的字符串匹配

dp[i]=min(dp[i+1]+1, dp[t]+t-i-m)，如果存在单词可以匹配，且recmsg[t-1]为可以匹配的字符串的最后一个字符，匹配的单词的长度为m。

 

要使dp[i]得到最小值，必须使dp[i+1]得到最小值。

 

因此根据这个递推式，可以得到dp[0]的最小值，即问题的最优化解。

 

#include <iostream> #include <string> using namespace std; int min(int a, int b) { return a>b? b:a; } int main() { int W,L; string recmsg, str; cin >> W >> L; cin >> recmsg; string *dict=new string[W]; int *dp=new int[L+1]; int i=0; int j=W; int k; while(j--) { cin >> str; dict[i++]=str; //W字典 } dp[L]=0; for(i=L-1;i>=0;i--) { dp[i]=dp[i+1]+1; for(j=0;j<W;j++) { if(dict[j].length()<=L-i && dict[j].at(0)==recmsg[i]) { int t=i; int m=0; while(t<L) { if(dict[j].at(m)==recmsg[t++]) m++; if(m==dict[j].length()) { dp[i]=min(dp[t]+t-i-m,dp[i]); break; } } } } } cout << dp[0] << endl; }

 

附录：

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3561		Accepted: 1564

Description

Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometimes they hear words that do not make any sense. For instance, Bessie once received a message that said "browndcodw". As it turns out, the intended message was "browncow" and the two letter "d"s were noise from other parts of the barnyard.

The cows want you to help them decipher a received message (also containing only characters in the range 'a'..'z') of length L (2 ≤ L ≤ 300) characters that is a bit garbled. In particular, they know that the message has some extra letters, and they want you to determine the smallest number of letters that must be removed to make the message a sequence of words from the dictionary.

Input

Line 1: Two space-separated integers, respectively: W and L
Line 2: L characters (followed by a newline, of course): the received message
Lines 3.. W+2: The cows' dictionary, one word per line

Output

Line 1: a single integer that is the smallest number of characters that need to be removed to make the message a sequence of dictionary words.

Sample Input

6 10
browndcodw
cow
milk
white
black
brown
farmer

Sample Output

2