列表

详情


NC25052. [USACO 2007 Feb S]The Cow Lexicon

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.

输入描述

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

输出描述

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.

示例1

输入:

6 10
browndcodw
cow
milk
white
black
brown
farmer

输出:

2

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

Python3(3.9) 解法, 执行用时: 558ms, 内存消耗: 4756K, 提交时间: 2022-04-04 22:31:26

d = {}
def search(sentence): 
    if sentence == '': return 0
    if sentence in d: return d[sentence]
    ans = len(sentence)
    for word in words:
        pos = 0
        pos_word = 0
        while pos < len(sentence):
            if sentence[pos] == word[pos_word]:
                pos_word += 1
                if pos_word == len(word): break
            pos += 1
        if pos_word == len(word):
            ans = min(ans, pos+1-len(word)+search(sentence[pos+1:]))
    d[sentence] = ans
    return ans

w, l = [int(x) for x in input().split()]
sentence = input()
words = set()
for _ in range(w):
    words.add(input())
print(search(sentence))

C++(g++ 7.5.0)(g++7.5.0) 解法, 执行用时: 90ms, 内存消耗: 384K, 提交时间: 2023-08-08 16:21:30

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[605][30], jud[305];
int dp[305];
int main(void)
{
	int n, m, i, j, p, q, sum;
	scanf("%d%d", &n, &m);
	scanf("%s", jud+1);
	for(i=1;i<=n;i++)
		scanf("%s", str[i]+1);
	for(i=0;i<=m;i++)
		dp[i] = i;
	for(i=1;i<=m;i++)
	{
		for(j=1;j<=n;j++)
		{
			sum = 0;
			p = i, q = strlen(str[j]+1);
			while(p>=1 && q>=1)
			{
				if(jud[p]==str[j][q])
					q--, sum++;
				p--;
			}
			if(q==0)
				dp[i] = min(dp[i], dp[p]+(i-p-(int)strlen(str[j]+1)));
		}
	}
	printf("%d\n", dp[m]);
	return 0;
}

C++(clang++11) 解法, 执行用时: 69ms, 内存消耗: 432K, 提交时间: 2022-04-13 16:55:46

#include<bits/stdc++.h>
using namespace std;
int n,m,f[1005];
char a[1005];
string s[1005];
int main()
{
	cin>>n>>m;
	for(int i=1;i<=m;i++)
	    cin>>a[i];
	for(int i=1;i<=n;i++) 
	    cin>>s[i];
	for(int i=1;i<=m;i++)
    {
		f[i]=f[i-1]+1;
	    for(int j=1;j<=n;j++)
	  	{
	  		int l=s[j].length()-1,l1=i;
	  		while(l>=0&&l1>=1)
            {
	  		    l-=(s[j][l]==a[l1]);
                l1--;
			}
			if(l==-1)
                f[i]=min(f[i],f[l1]+i-l1-int(s[j].length()));
		}
	}
	cout<<f[m];
	return 0;
}

上一题