POJ3267-The Cow Lexicon(dp)

POJ3276 原题链接:http://poj.org/problem?id=3267

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10896		Accepted: 5214

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

题目大意:给一串字符串,求最少删掉几个字母可以与下面的单词匹配;

输入：

6 10(一共有6个单词,字符串的长度为10)

browndcodw(长度为10的字符串)

cow

mile

white

black

brown

farmer  (6个单词)

思路:

反向(从L到0,所求为dp[0]):

1.dp,dp[i]表示从i到L所删除的最小字母数;

2.首先明确,若字符串与下面的单词无匹配,那么dp[i]=dp[i+1]+1;

3.若匹配,则dp[i]=min(dp[i],dp[pm]+pm-i-len)

   (1).这里pm是字符串的指针（其中i表示当前所匹配的单词在字符串中的起始位置）pd是单词的指针,len是单词的长度;

   (2).pm-i-len表意为:从i到pm所删除的字母数,dp[pm]表示为从pm到L所删除的字符数,所以dp[pm]+pm-i-len表示i到L所删除的字符数;

   (3).因为删除方式不止一种,在i到L的区间内与字符串匹配的单词也不一定只一个所以要将dp[i]与dp[pm]+pm-i-len比较取最小值;

4.dp[0]即为所求,3步请结合程序观看;

代码如下 :

#include<iostream>
#include<string>
using namespace std;
int min(int a,int b)
{
    return a<b?a:b;
}
int main()
{
    int i,j;
    int w,L;
    while(cin>>w>>L)
    {
        int* dp=new int[L+1];
        char* xx=new char[L];
        string* dict=new string[w];

        cin>>xx;
        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;  //无匹配时的情况即步骤1
            for(j=0;j<w;j++) //枚举所有单词
            {
                int len=dict[j].length();
                if(len<=L-i && dict[j][0]==xx[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;  //字符串的指针
                    int pd=0;  //单词的指针
                    while(pm<L) //单词逐字匹配
                    {
                        if(dict[j][pd]==xx[pm++])
                            pd++;
                        if(pd==len)
                        {     //字典单词和message可以匹配时，状态优化（更新）
                            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删除的字符数
                    }
                }
            }
        }
        for(int i=0;i<L;i++)
        {
            cout<<dp[i]<<" ";
        }
        delete dp,xx,dict;  //释放空间
    }
    return 0;
}