zoukankan      html  css  js  c++  java
  • POJ 3267, The Cow Lexicon

    Time Limit: 2000MS  Memory Limit: 65536K
    Total Submissions: 2856  Accepted: 1252


    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


    // POJ3267.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <string>
    using namespace std;

    int main(int argc, char* argv[])
    {
        
    char words[600][26];
        
    char inw[301];
        
    int W, L;

        scanf(
    "%d %d\n",&W,&L);
        gets(inw);
        
    for (int i = 0; i < W; ++i)gets(words[i]);

        
    int del[301];
        del[L] 
    = 0;
        
    for(int i = L - 1 ; i >= 0 ; --i)
        {
            del[i] 
    = 301;
            
    for(int j = 0 ; j < W ; j ++)
            {
                
    int k = 0,l = i;
                
    for(;words[j][k] && l < L; ++l)
                    
    if(words[j][k] == inw[l])k++;

                
    if(!words[j][k])
                    del[i] 
    = std::min(std::min(del[i],del[i + 1+ 1),del[l] + l - i - k);
                
    else
                    del[i] 
    = min(del[i], del[i + 1+ 1);
            }
        }
        cout
    <<del[0]<<endl;
        
    return 0;
    }

  • 相关阅读:
    Android开发 View_自定义快速索引侧边栏 SideBarView
    Android开发 PopupWindow开发的一些例子
    Android开发 Fragment里监听返回键
    AndroidStudio 清除项目里无用的资源
    Android开发 SingleLiveEvent解决LiveData或者MutableLiveData多次回调的问题
    字母排列城市列表资源
    Android开发 GridView详解
    Android开发 NestedScrollView嵌套RecyclerView导致的UI加载慢的问题
    Android开发 跳转指定应用商城评分
    Android开发 Activity生命周期详解
  • 原文地址:https://www.cnblogs.com/asuran/p/1582222.html
Copyright © 2011-2022 走看看