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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8458 | Accepted: 3993 |
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
Output
Sample Input
6 10 browndcodw cow milk white black brown farmer
Sample Output
2
题意：删掉给定单词的部分字母使之能由词典中的单词组合而成，求删掉的最少字母数量
思路：dp，dp[i]=min{dp[i+1]+1,dp[ps]+(i-ps-len)}，dp[i]表示从第i位起的后缀中删掉的最少字母数量使该后缀能由词典中的单词合成，答案即为dp[0]
#include<iostream> #include<cstring> #include<string> #include<algorithm> using namespace std; const int maxn=1010; int W,L; string str,dict[maxn]; int dp[maxn]; int main() { while(cin>>W>>L){ cin>>str; for(int i=0;i<W;i++) cin>>dict[i]; dp[L]=0; for(int i=L-1;i>=0;i--){ dp[i]=dp[i+1]+1; for(int k=0;k<W;k++){ int len=dict[k].length(); if(dict[k][0]!=str[i]) continue; int ps=i,pd=0; while(ps<L&&pd<len&&L-ps>=len-pd){ if(dict[k][pd]==str[ps]) pd++; ps++; } if(pd==len) dp[i]=min(dp[i],dp[ps]+(ps-i-len)); } } cout<<dp[0]<<endl; } return 0; }
2015-03-19 22:20:21
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原文地址：http://www.cnblogs.com/--560/p/4351955.html
