poj3267

The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 4716 Accepted: 2144 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

 

DP: http://blog.csdn.net/AllenLSY/archive/2009/12/08/4966796.aspx

此题居然能1Y，我很是欣喜

AC后，看了一下网上的解法，没有特别详细的，我提供一个。

网上的讨论，有人说从后面往前面DP会快一些，我觉得没有道理吧。从前往后与从后往前，在DP最优值的递推上要相反，除此以外求remove值时对单词的扫描也要相反（后面会详细解释），基本原理是一样的，复杂度应该是没有区别的。

设d[i:0..l-1]表示从头到i位（i位包含）最少需要除去多少个字符，S是原串。

d[i]=min( d[j]+remove //如果S[j+1..i]子串中包含了一个单词，j<i
                d[i-1]+1       //如果不存在可包含单词的子串 )

因此，主要步骤变成了如何求remove

假如我们把单词都存在w里面，对于其中的一个单词w[k]，让now=len[k]，j=i开始，找一个j>=0，使得S[j..i]之间包含单词w[k]。所谓包含，其实也就是说w[k]与S[j..i]的LCS是w[k]，不过这里我们不用LCS算法，因为慢。接着刚才说，如果S[j..i]之间不包含单词w[k]，则remove=i。如果包含，则remove=i-j+1-len[k]。

在我的程序里，这里还有一个优化，如果求remove的循环过程中，i-j+1-len[k]<d[i](前面求出的d[i])，则终止继续求remove，j停止--，因为此时即使求出j也会比原d[i]大了，所以让remove=i-j就可以。

#include <iostream>
#include <cstring>

using namespace std;

int main()
{
    char dic[600][26];
    int W,L;
    char msg[301];
    int stat[301];
    int del, mindel;
    int mstart, dstart;
    int diclen;

    while(cin>>W>>L>>msg)
    {

        for(int i=0; i<W; i++)
        {
            cin>>dic[i];
        }

        memset(stat,0,sizeof(stat));
        for(int i=L-1; i>=0; i--)
        {
            mindel=601;
            for(int j=0; j<W; j++)
            {
                if(msg[i]!=dic[j][0])
                {
                    continue;
                }
                del=0;//match
                mstart=i+1;
                dstart=1;
                diclen=strlen(dic[j]);
                while(mstart<L&&dstart<diclen)
                {
                    if(msg[mstart]!=dic[j][dstart])
                    {
                        del++;
                    }
                    else
                    {
                        dstart++;
                    }
                    mstart++;
                }
                if(dstart>=diclen)//match
                {
                    mindel=del+stat[mstart]>mindel? mindel: del+stat[mstart];
                    //cout<<"i:"<<i<<"j:"<<j<<endl;
                    //cout<<"mindel:"<<mindel<<endl;
                }
            }
            stat[i]=mindel<stat[i+1]+1? mindel: stat[i+1]+1;


        }
        cout<<stat[0]<<endl;

    }

    return 0;
}