POJ 3267 The Cow Lexicon(动态规划)

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5719		Accepted: 2638

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

 解题报告：题意就是给出一个序列，和其他的单词进行匹配，求最少去掉的单词数，思路参考的网上的，代码中有解释；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 610;
int len[N], dp[N];//dp[i]表示第i个字符串最少删除的字符数
char str[N], word[N][N];
int L, W;
int Min(int a,int b)
{
    return a < b ? a : b;
}
void DP()
{
    int i, j, k, p;
    memset(dp, 0, sizeof(dp));//初始化
    for(i = 1; i <= L; ++i)//循环单词的长度，
    {
        dp[i] = dp[i - 1] + 1;
        for (j = 1; j <= W; ++j)
        {
            k = len[j] - 1;//所要匹配的单词的最后字母的位置
            p = i - 1;
            while (k >= 0 && p >= 0)//从后向前匹配
            {
                if (word[j][k] == str[p])
                {
                    k --;
                }
                p --;
            }
            if (k < 0)//若k<0说明匹配成功时
            {
                dp[i] = Min(dp[i], dp[p + 1] + i - p - 1 - len[j]);
                //状态方程（i - p - 1 -len[j]）第j个单词匹配成功时多余的单词数，列如browndcodw
                //和cow匹配时，从后向前匹配此时i = 10；匹配成功时正好匹配到codw,四个单词，而cow是三个，
                //所以（i - p - 1 -len[j]）= 1；
            }
        }
    }
}
int main()
{
    int i;
    scanf("%d%d", &W, &L);
    scanf("%s", str);
    memset(word, 0, sizeof(word));
    for (i = 1; i <= W; ++i)
    {
        scanf("%s", word[i]);
        len[i] = strlen(word[i]);
    }
    DP();
    printf("%d\n", dp[L]);
    return 0;
}