[USACO2007FEB S] The Cow Lexicon S

题目描述

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.

输入格式

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

输出格式

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.

样例 #1

样例输入 #1

6 10
browndcodw
cow
milk
white
black
brown
farmer

样例输出 #1

2

首先考虑暴力。设计\(dp_i\)表示前i个字符要去掉多少个字符可以被翻译。暴力转移就是枚举上一个单词的结尾在j，在j+1~i中只能有一个单词，那么价值是\((i-j)-\)最长的单词。不妨枚举选了哪一种单词，然后看一下能不能匹配。

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
int w,l,dp[N],len[N];
char s[N][N],t[N];
int can(int x,int y)
{
	int ans=y-x+1;
	for(int k=1;k<=w;k++)//枚举选哪一种单词
	{
		int r=1;
		for(int i=x;i<=y;i++)
		{
			if(s[k][r]==t[i])
				++r;
			if(r>len[k])//可以匹配上
				ans=min(ans,y-x+1-len[k]);
		}
	}
	return ans;
}
int main()
{
	scanf("%d%d%s",&w,&l,t+1);
	for(int i=0;i<=l;i++)
		dp[i]=i;
	for(int i=1;i<=w;i++)
		scanf("%s",s[i]+1),len[i]=strlen(s[i]+1);
	for(int i=1;i<=l;i++)
		for(int j=0;j<i;j++)
			dp[i]=min(dp[i],dp[j]+can(j+1,i));
	printf("%d",dp[l]);
	return 0;
}

但是会超时获得70分。我们发现可以预处理出每个can(l,r)。枚举l和选哪种单词，然后往后推r，如果一个单词在某一处可以被匹配那么在后面都可以匹配的上。

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
int w,l,dp[N],len[N],f[N][N];
char s[N][N],t[N];
int can(int x,int y)
{
	int ans=2e9;
	for(int k=1;k<=w;k++)
	{
		int r=1;
		for(int i=x;i<=y;i++)
		{
			if(s[k][r]==t[i])
				++r;
			if(r>len[k])
				ans=min(ans,y-x+1-len[k]);
		}
	}
	return ans;
}
int main()
{
	memset(f,0x7f,sizeof(f));
	scanf("%d%d%s",&w,&l,t+1);
	for(int i=0;i<=l;i++)
		dp[i]=i;
	for(int i=1;i<=w;i++)
		scanf("%s",s[i]+1),len[i]=strlen(s[i]+1);
//	return can(6,10);
	for(int i=1;i<=l;i++)
	{
		int ans=2e9;
		for(int k=1;k<=w;k++)
		{
			int r=1;
			for(int j=i;j<=l;j++)
			{
				if(s[k][r]==t[j])
					++r;
				if(r>len[k])
					f[i][j]=min(f[i][j],j-i+1-len[k]);
			}
		}
	}
	for(int i=1;i<=l;i++)
		for(int j=0;j<i;j++)
			dp[i]=min(dp[i],dp[j]+f[j+1][i]);
	printf("%d",dp[l]);
	return 0;
}