The Cow Lexicon
| Time Limit: 2000MS |
|
Memory Limit: 65536K |
| Total Submissions: 5619 |
|
Accepted: 2599 |
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
这个题目还是跟最长公共子序列类似，
状态表示还是用f[i]表示前i个中最少去掉的数目能够构成合法序列
则f[i]=i
f[i]=min(f[i],f[j]+remove(s[j+1..i],ss[k]))
if ss[k]能在s[j+1..i]中表示出来
所以我们不必用LCS来求这个remove，直接用pp[k]表示第k个单词能在s[j+1..i]中匹配到的地方的地方
如果单词能在其中表示出来，则min
这类题目的类似在于f[i]总是由f[j](0<j<i)推出来的
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 350
5int f[MAX];
6char sd[MAX];
7int pp[610],len[610];
8char s[610][26];
9int l1,n,i,j,k;
10int min(int a,int b)
11
{
12
if (a>b) return b;
13 else return a;
14
}
15int main()
16{
17 scanf("%d%d",&n,&l1);
18 scanf("%s",&sd);
19 for (i=l1-1;i>=0 ;i-- )
20
{
21 sd[i+1]=sd[i];
22
}
23 for (i=1; i<=n ; i++ )
24 {
25 scanf("%s",&s[i]);
26 len[i]=strlen(s[i]);
27 }
28 for (i=1; i<=l1 ; i++ )
29 {
30 f[i]=i;
31 for (j=1; j<=n ; j++)
32 {
33 pp[j]=len[j]-1;
34 }
35 for (j=i; j>=1; j--)
36 {
37 for (k=1; k<=n ; k++ )
38 {
39 if (sd[j]==s[k][pp[k]])
40 {
41 pp[k]--;
42 }
43 if (pp[k]<0)
44 {
45 f[i]=min(f[i],f[j-1]+i-j-len[k]+1);
46 }
47 }
48 }
49 }
50 printf("%d\n",f[l1]);
51 return 0;
52}
53//f[i]=min(f[j]+remove(s[j+1..i],s[k]))
54//
55