（POJ3267）The Cow Lexicon <DP>

The Cow Lexicon
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

题意：
给你一个长度为L的字符串，和W个长度<= 25 的单词。问你最少去掉多少个字母可以使剩下的字符串全部由单词组成？

分析：
设dp[i] 表示将字符串[i,L)的字母变为单词最少要去掉的字母数。
初始条件：dp[L] = 0;
状态转移方程： dp[i] = min(dp[i+1] + 1, dp[t] + t - i - m);
(1):当mes[0] 和所有的单词首字母都不相同时，则dp[i] = dp[i+1] + 1;
(2):判断mes[i,L)中是否包含某个单词（不一定要连续，可以间隔包含）。若某个单词最后一个字母匹配的位置为t-1,那么dp[i] = dp[t] + t - i - m;
如： 单词为cown ,mes=”cowdn” 时 最后匹配的位置为t=5,m=4,所以dp[0] = dp[5] + 5 - 0 - 4 = 1;

函数：string.at(i) 返回第i个字母
string *ws = new string[n];创建大小为n的string数组

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <vector>using namespace std;int L,W;string mes;int dp[310];int main(){    string *ws = new string[610];//创建字符串数组    while(scanf("%d%d",&W,&L)!=EOF)    {        cin>>mes;        for(int i=0;i<W;i++)            cin>>ws[i];        dp[L] = 0;        for(int i=L-1;i>=0;i--)        {            //默认开头字母不匹配            dp[i] = dp[i+1] + 1;            //            for(int j=0;j<W;j++)            {                if(L-i>=ws[j].length() && mes[i] == ws[j].at(0))                {                    int t = i;                    int m = 0;                    while(t < L)                    {                        if(ws[j].at(m) == mes[t++])  m++;//注意：当==号不成立时t++，而m不++                        if(m == ws[j].length())//当匹配了m个字母后，但不一定是连续匹配的                        {                            dp[i] = min(dp[i],dp[t]+t-i-m);                            break;                        }                    }                }            }        }        printf("%d\n",dp[0]);    }    return 0;}