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POJ 3267 The Cow Lexicon
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10021 | Accepted: 4814 |
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 2: L characters (followed by a newline, of course): the received message
Lines 3..W+2: The cows‘ dictionary, one word per line
Output
Sample Input
6 10browndcodwcowmilkwhiteblackbrownfarmerSample Output
2Source
#include<iostream>#include<cstdio>#include<cstring>using namespace std;int w,l;char ch[610];string vec[610];int f[610];int main(){ cin>>w>>l; for(int i=0;i<l;i++) cin>>ch[i];// 字符串 for(int i=0;i<w;i++) cin>>vec[i];// 多个单词 f[l]=0; for(int i=l-1;i>=0;i--) //对于字符串 我采取 从后向前更新的 方法 { f[i]=f[i+1]+1;// 最坏情况 当前字符需要删除 固由 f[i+1]+1得来 for(int j=0;j<w;j++)// 遍历每个单词 { int len=vec[j].length();// 获取单词的长度 if(len<=l-i+1&&vec[j][0]==ch[i]) // 当前遍历到的 字符串 中的 字符到字符串尾 的长度 大于等于 当前查询的单词的长度 {// 切当前遍历到的字符串中的字符 等于 该单词的首字母 int pm=i;// 字符串的指针 int pz=0;// 单词的指针 while(pm<l)// 检索从此 向后 是否逐字匹配 { if(vec[j][pz]==ch[pm++]) pz++; // 继续向后检索 if(pz==len) { f[i]=min(f[i],f[pm]+(pm-i)-len);break;// 检所完成 从i 向后到l与该单词逐字匹配 所以 更新维护f[] } } } } } cout<<f[0]; return 0;}
慢慢推 DP f[pm]+(pm-i)-len 这个地方是我从网上找的 至今没搞明白
POJ 3267 The Cow Lexicon