POJ 3267 The Cow Lexicon

最新推荐文章于 2021-07-12 12:53:54 发布

原创最新推荐文章于 2021-07-12 12:53:54 发布 · 561 阅读

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本文探讨了一种解决牛群间通信中噪声干扰的问题，即确定收到消息中需要删除的最少字符数量，使其成为字典词汇序列。通过实例分析，详细解释了解决该问题的算法步骤。

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The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6548		Accepted: 3021

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

Source

USACO 2007 February Silver

做这个题目的时候是一点思路没有，感觉用dp但是不会搞，所以直接看了一下解题报告

http://blog.youkuaiyun.com/lyy289065406/article/details/6648121 感觉这个挺好的

#include <iostream>
#include <string.h>
using namespace std;
int dp[400];
string s1[610];
string s2;
int min1;
int main()
{
    int i,j,n,m,s,t;
    int len,l,x,y;
    cin>>n>>len;
    cin>>s2;
    for(i=1; i<=n; i++)
    {
        cin>>s1[i];
    }
    dp[len]=0;
    for(i=len-1; i>=0; i--)
    {
        dp[i]=dp[i+1]+1;
        for(j=1; j<=n; j++)
        {
            l=s1[j].size();
            if(l<=(len-i)&&s2[i]==s1[j][0])
            {
                for(x=i,y=0; x<=len-1; x++)
                {
                    if(s2[x]==s1[j][y])
                    {
                        y++;
                    }
                    if(y==l)
                    {
                        if(dp[i]>(dp[x+1]+(x+1)-i-l))
                        {
                            dp[i]=(dp[x+1]+(x+1)-i-l);
                        }
                    }
                }
            }
        }
    }
    cout<<dp[0]<<endl;
    return 0;
}