POJ 3267 The Cow Lexicon

The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9759 Accepted: 4678 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 Source USACO 2007 February Silver

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　　这个题半天没读懂，看了半天评论后大概是这么个意思：第一行输入一个长度为Ｌ的字母串，剩下的Ｗ行为Ｗ个字典单词，然后让字符串与每一个字典单词进行匹配，问字符串最少需要删除的字母的个数。这道题就是ＤＰ＋字符串匹配了，动态转移方程如下：

动态转移方程: DP[i]=DP[i+1]+1;(没有满足匹配的单词)

　　　　　　 DP[i]=min(DP[i],DP[pm]+(pm-i)+len);(有满足匹配的单词)

　　DP[i]表示的是message[i]到message[len-1]的满足的最小出去字母的个数，如果字典单词中没有能够匹配的，ＤＰ[i]直接等于ＤＰ[i+1]+1 ；如果存在能够匹配的单词，DP[i]=min(DP[i],DP[pm]+(pm-i)+len)，这个就拿第一个为例：

　　message: browndcodw -----> browndcodw

dir[0]: cow co w

此时ｐｍ(ｍｅｓｓａｇｅ的指针)数值为１０，ｐｄ（ｄｉｒ的指针）数值为３，这期间没有匹配的字母（这里就为ｄ）的个数为(pm-i)+len（ｉ为ｍｅｓｓａｇｅ初始指针初始位置），加上第ｐｍ之后需要删除的最小值（这里是ｄｐ[10],数值为零）总的就为　DP[i]=min(DP[i],DP[pm]+(pm-i)+len);

//880K 329MS

#include <iostream>

#include <cstring>

#include <algorithm>

using namespace std;

int dp[300+5];

int main()

{

char message[300+5], dir[600+5][300+5];

int w, l;

cin>>w>>l;

cin>>message;

for(int i=0;i<w;i++)

cin>>dir[i];

dp[l]=0;

for(int i=l-1;i>=0;i--) {

dp[i]=dp[i+1]+1;

for(int j=0;j<w;j++) {

int pm=i, pd=0;

int len=strlen(dir[j]);

if(len<=l-i) {

while(pm<l) {

if(message[pm++]==dir[j][pd])

pd++;

if(pd==len) {cout<<i<<' '<<j<<' '<<pm<<' '<<i<<' '<<len<<endl;dp[i]=min(dp[i],dp[pm]+(pm-i)-len);break;}

}

}

}

}

cout<<dp[0]<<endl;

return 0;

}