The Cow Lexicon
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.
//本题具有子最优结构,适合运用动态规划,利用数组dp[]保存已经计算过的数值,当再次用到时直接调用 //题意 从给定一个字符串和若干个单词,考虑从字符串中至少删除多少个字符,可使字符串与单词匹配 #include<iostream> #include<string> using namespace std; int main() { int i ,j; int w,L; while(cin>>w>>L) { int* dp=new int[L+1]; char* mesg=new char[L]; string* dict=new string[w]; cin>>mesg; for(i=0;i<w;i++) cin>>dict[i]; dp[L]=0; //dp[i]表示从i到L所删除的字符数 for(i=L-1;i>=0;i--) //从message尾部开始向前检索 { dp[i]=dp[i+1]+1; //字典单词和message无法匹配时,删除的字符数(最坏的情况) for(j=0;j<w;j++) //对字典单词枚举 { int len=dict[j].length(); if(len<=L-i && dict[j][0]==mesg[i]) //单词长度小于等于当前待匹配message长度 { //且单词头字母与信息第i个字母相同 int pm=i; //message的指针 int pd=0; //单词的指针 while(pm<L) //单词逐字匹配 { if(dict[j][pd]==mesg[pm]) { pd++; pm++; } else pm++; if(pd==len) { //dp[i]中原来保存的是 从i到L 无匹配单词时的需要删除的字符数 dp[i]=min(dp[i],dp[pm]+(pm-i)-len);//dp[pm]表示从pm到L删除的字符数 break; //(pm-i)-pd表示从i到pm删除的字符数 } //则dp[pm]+(pm-i)-pd表示从i到L有匹配单词时需要删除的字符数 } } } } cout<<dp[0]<<endl; //表示从第一个单词到最后一个单词这个区间需要删除的最少字符数 } return 0; }