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  • poj3267——线性dp

    poj3267——线性dp

    The Cow Lexicon
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 8458   Accepted: 3993

    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

    题意:删掉给定单词的部分字母使之能由词典中的单词组合而成,求删掉的最少字母数量
    思路:dp,dp[i]=min{dp[i+1]+1,dp[ps]+(i-ps-len)},dp[i]表示从第i位起的后缀中删掉的最少字母数量使该后缀能由词典中的单词合成,答案即为dp[0]
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    
    using namespace std;
    
    const int maxn=1010;
    int W,L;
    string str,dict[maxn];
    int dp[maxn];
    
    int main()
    {
        while(cin>>W>>L){
            cin>>str;
            for(int i=0;i<W;i++) cin>>dict[i];
            dp[L]=0;
            for(int i=L-1;i>=0;i--){
                dp[i]=dp[i+1]+1;
                for(int k=0;k<W;k++){
                    int len=dict[k].length();
                    if(dict[k][0]!=str[i]) continue;
                    int ps=i,pd=0;
                    while(ps<L&&pd<len&&L-ps>=len-pd){
                        if(dict[k][pd]==str[ps]) pd++;
                        ps++;
                    }
                    if(pd==len) dp[i]=min(dp[i],dp[ps]+(ps-i-len));
                }
            }
            cout<<dp[0]<<endl;
        }
        return 0;
    }
    View Code

    2015-03-19 22:20:21

    没有AC不了的题,只有不努力的ACMER!
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  • 原文地址:https://www.cnblogs.com/--560/p/4351955.html
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