poj 3267 The Cow Lexicon (动态规划)

最新推荐文章于 2020-05-20 22:51:32 发布

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最新推荐文章于 2020-05-20 22:51:32 发布

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文章标签：数据结构与算法

本文介绍了一种算法，用于解析牛群的通讯信息。通过反转字典中的词汇并使用动态规划的方法，找到最少需要删除的字符数，使接收到的混乱信息能够组成有意义的词汇序列。

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The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8167		Accepted: 3845

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

dp[i]代表以i为结尾的最优选择,状态转移方程为：

dp[i]=min(dp[i-1]+1,opt); opt:包括第 i个字符的最优选择；

把原先的字典里的词组反转，枚举字典里的全部词汇，找到最优解。

#include<stdio.h>
#include<queue>
#include<map>
#include<string>
#include<string.h>
using namespace std;
#define N 305
const int inf=0x1f1f1f1f;
char str[N*2][30],s[N];
int dp[N];
int main()
{
    int i,j,k,n,m;
    char ch;
    while(scanf("%d%d",&m,&n)!=-1)
    {
        scanf("%s",s+1);
        for(i=0;i<m;i++)
        {
            scanf("%s",str[i]);
            int len=strlen(str[i]);  //字符串反转
            for(j=0;j<len/2;j++)
            {
                ch=str[i][j];
                str[i][j]=str[i][len-j-1];
                str[i][len-1-j]=ch;
            }
            str[i][len]='\0';
        }
        dp[0]=0;
        int tmp;
        for(i=1;i<=n;i++)
        {
            tmp=dp[i-1]+1;  //tmp初始化为该字符舍去时的值
            for(j=0;j<m;j++)
            {
                if(str[j][0]!=s[i])      
                    continue;
                int l=1,len=strlen(str[j]);
                for(k=i-1;k>0&&l<len;k--)  //寻找该单词出现的最早位置
                {
                    if(s[k]==str[j][l])
                        l++;
                }
                if(l==len)      //包括此单词
                    tmp=min(tmp,dp[k]+(i-k-len));
            }
            dp[i]=tmp;
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}