传送门 f[i] 表示前 i 个字符去掉多少个 的最优解 直接暴力DP ——代码 #include <cstdio> #include <cstring> #include <iostream> ]; ], a[][]; inline int read() { , f = ; char ch = getchar(); ; ) + (x << ) + ch - '; return x * f; } inline int min(int x, int y) {…

P2875 [USACO07FEB]牛的词汇The Cow Lexicon 三维dp 它慢,但它好写. 直接根据题意设三个状态: $f[i][j][k]$表示主串扫到第$i$个字母,匹配到第$j$个单词的第$k$位可以留下的最多字符数 当该位不选时,就传递上一位的数据$f[i][j][k]=f[i-1][j][k]$ 当该位可以匹配时: $if(a[i]==b[j][k]\&\&f[i-1][j][k-1])$$f[i][j][k]=max(f[i][j][k],f[i-1][j][k-1…

P2875 [USACO07FEB]牛的词汇The Cow Lexicon 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; sometime…

https://daniu.luogu.org/problemnew/show/P2875 dp[i]表示前i-1个字符,最少删除多少个 枚举位置i, 如果打算从i开始匹配, 枚举单词j,计算从i开始往后,匹配完单词j的位置pos,删除的字母个数sum dp[pos]=min(dp[i]+sum) 如果不用i匹配,dp[i+1]=min(dp[i]) #include<cstdio> #include<cstring> #include<iostream> using…

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…

The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9041   Accepted: 4293 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 c…

The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8211   Accepted: 3864 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 c…

The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8252   Accepted: 3888 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 c…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7290   Accepted: 3409 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 sys…

题目大意: 输入w,l: w是接下来的字典内的单词个数,l为目标字符串长度 输入目标字符串 接下来w行,输入字典内的各个单词 输出目标字符串最少删除多少个字母就能变成只由字典内的单词组成的字符串 Sample Input 6 10browndcodwcowmilkwhiteblackbrownfarmer Sample Output 2   当输入为 2 6 reader rad reder 输出为 1 即单词不能相互覆盖 只能各自独立组合 #include <algorithm> #incl…