POJ3267The Cow Lexicon(AC)

The Cow Lexicon

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10852 Accepted: 5197

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 10browndcodwcowmilkwhiteblackbrownfarmer

Sample Output

2

//牛也有自己的语言，但是有时候会有多余的字母，因为需要去掉最少的字母，然后找到需要的单词//动态规划不是自己擅长的题目//这个应该也是要用到字符串匹配了，建个树吧//动态规划自己不擅长，主要是状态方程应该怎么定//2种方法从前往后和从后往前都AC#define _CRT_SECURE_NO_WARNINGS#include <stdio.h>#include <malloc.h>int W = 0;int L = 0;#define MAXL 305#define MAXW 605char word[MAXL];char dictionary[MAXW][MAXL];typedef struct node{char word[MAXL];int isword;node* next[26];}nodes;nodes* trees;nodes* head = NULL;int dp[MAXL];void init(){int i = 0;int j = 0;for (i = 0; i < MAXW;i++){for (j = 0; j < MAXL; j++){dictionary[i][j] = '\0';}}for (i = 0; i < MAXL; i++){word[i] = '\0';dp[i] = 0;}return;}int strlength(char* s){int size = 0;while ('\0'!= *s){s += 1;size += 1;}return size;}void maketree(int index){nodes* p = NULL;nodes* tmp = NULL;int size = 0;int i = 0;int j = 0;int nextword = 0;size = strlength(dictionary[index]);p = head;for (i = 0; i < size;i++){nextword = dictionary[index][i] - 'a';if (NULL == p->next[nextword])//没有过这个字母，那就要重新创建{p->next[nextword] = (nodes*)malloc(sizeof(nodes));if (i == size-1){p->next[nextword]->isword = 1;for (j = 0; j < size; j++){p->next[nextword]->word[j] = dictionary[index][j];}p->next[nextword]->word[size] = '\0';}else{p->next[nextword]->isword = 0;p->next[nextword]->word[0] = '\0';}for (j = 0; j < 26; j++){p->next[nextword]->next[j] = NULL;}}else{if (i == size - 1){p->next[nextword]->isword = 1;for (j = 0; j < size; j++){p->next[nextword]->word[j] = dictionary[index][j];}p->next[nextword]->word[size] = '\0';}}p = p->next[nextword];}return;}int min(int a, int b ){if (a<=b){return a;}else{return b;}}int main(){int i = 0;int j = 0;int k = 0;int len = 0;int pw = 0; //输入单词的位置int pd = 0; //字典单词的位置init();freopen("input.txt","r",stdin);scanf("%d %d",&W,&L);scanf("%s",word);//先创建一个根节点head =(nodes*) malloc(sizeof(nodes));head->word[0] = '\0';head->isword = 0;for (i = 0; i < 26;i++){head->next[i] = NULL;}for (i = 1; i <= W;i++){scanf("%s",dictionary[i]);//maketree(i); //字典树建好了}//动态规划，这个状态方程好难啊。。。//从别人的解题报告中得到的思路，自己不会推状态方程//dp[i]表示从s的第0个字符开始长度为i的字符串(末位为i-1，后面的字符不考虑)需要删除的字符数//貌似没用到字典树，因为单词长度比较短//为什么要从尾部开始比，我比较奇怪//我自己决定从前往后不是更好理解吗？//从尾部找起/*for (i = L - 1; i >= 0;i--){dp[i] = dp[i+1] + 1;  //最差的情况是新的序列多的字母删掉for (j = 1; j <= W;j++) //开始遍历字典{len = strlength(dictionary[j]); //长度//挨个比较字符是否匹配//查看是否有字符串是匹配的，当然中间可能是有些字符是多余的那就要删掉if (len<=(L-i) &&(dictionary[j][0] == word[i])) //长度小于等于我们现在新增的，然后字典那个单词要一致{//这个比较的地方自己也是有点犹豫了pw = i; //从i开始一直到L-1pd = 0;while (pw<L){//找到匹配的单词，那字典就可以移一位if (dictionary[j][pd] == word[pw]){pd += 1;}pw+=1;if (pd == len) //找到单词匹配{dp[i] = min(dp[i], (dp[pw] + pw-i-len)); //pw-i-len表示 pw -i 表示移动了多长的字符串，然后匹配到len的字符，因为中间可能会有多的字符break;}}}}}printf("%d\n", dp[0]); //dp[0]表示0~L-1的最小删除字符串数目*///另一种方法，从前往后找 AC//dp[0] 表示0到0，dp[1]表示0~1,最后的的结果就是dp[L-1]for (i = 0; i <= L - 1;i++){//如果是从头往后，那第一个dp[0]的值最坏定为1if (0 == i){dp[i] = 1;}else{dp[i] = dp[i-1]+1;}//找匹配的字典单词for (j = 1; j <= W; j++) //开始遍历字典{len = strlength(dictionary[j]); //长度//挨个比较字符是否匹配//查看是否有字符串是匹配的，当然中间可能是有些字符是多余的那就要删掉//是的，从前往后的话，那比较单词就要从最后一个单词比较起了if (len <= (i+1) && (dictionary[j][len-1] == word[i])) //长度小于等于我们现在新增的，然后字典那个单词要一致{//这个比较的地方自己也是有点犹豫了pw = i; //从i开始一直到L-1pd = len-1;while (pw>=0){//找到匹配的单词，那字典就可以移一位if (dictionary[j][pd] == word[pw]){pd -= 1;}pw -= 1;if (pd == -1) //找到单词匹配{if (-1 == pw){dp[pw] = 0; //如果是正好前3个单词就是一个字典单词，那这个时候pw就=-1}dp[i] = min(dp[i], (dp[pw] + i - pw - len)); //i - pw -len表示 i-pw 表示移动了多长的字符串,因为从0开始的，然后匹配到len的字符，因为中间可能会有多的字符break;}}}}}printf("%d\n", dp[L-1]);return 0;}