题意  给出一个母串  和一个字典 问母串最少删去几个字母     删去后的母串是由字典里面的单词拼起来的 思路:dp[i]表示从i到母串结尾最少需要删除多少个字母  初始化dp[length]=0 最坏情况dp[i]=dp[i+1]+1 状态转移方程 dp[i]=min(dp[i],dp[p]+p-len-i) p 匹配单词的最后一个位置  len是字典里面单词的长度 i是开始匹配的位置   p-len-i的意义就是匹配过程中删除了多少个字母 参考:http://www.cnblogs.com…

http://poj.org/problem?id=3267 The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9380   Accepted: 4469 Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of t…

又见面了,还是原来的配方,还是熟悉的DP....直接秒了... The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7316 Accepted: 3421 Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the…

The Cow Lexicon Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8167   Accepted: 3845 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: 8905   Accepted: 4228 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…

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: 8815 Accepted: 4162 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 cowmu…

The Cow Lexicon Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 16   Accepted Submission(s) : 10 Problem Description Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words,…

The Cow Lexicon DescriptionFew 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 word…

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…

题目 1633: [Usaco2007 Feb]The Cow Lexicon 牛的词典 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 401  Solved: 216[Submit][Status] Description 没有几个人知道,奶牛有她们自己的字典,里面的有W (1 ≤ W ≤ 600)个词,每个词的长度不超过25,且由小写字母组成.她们在交流时,由于各种原因,用词总是不那么准确.比如,贝茜听到有人对她说"browndcodw"…

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…

The Cow Lexicon Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 11846 Accepted: 5693 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 cowmu…

题目:http://poj.org/problem?id=3267 题意:给定一个字符串,又给n个单词,求最少删除字符串里几个字母,能匹配到n个单词里 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cm…

题目链接: http://poj.org/problem?id=3267 从后往前遍历,dp[i]表示第i个字符到最后一个字符删除的字符个数. 状态转移方程为: dp[i] = dp[i+1] + 1;                                                 //当不能匹配时 dp[i] = std::min(dp[i], dp[msg] + (msg-i) - len[j]);  //当匹配时. 第i个字符到第msg个字符之间一共有msg-i个字符,减去…

https://vjudge.net/problem/POJ-3267 题意 给一个长度为L的字符串,以及有W个单词的词典.问最少需要从主串中删除几个字母,使其可以由词典的单词组成. 分析 状态设置很关键,设dp[i]表示以i为起始的后缀需要删去字母的最小数目.那么根据状态,必须从后往前遍历,现在考虑往前加入一个新字母,会发生什么呢?第一,考虑最坏情况,就是加进来的字母没有用处,即转移成dp[i+1]+1;第二,就是加入这个字母后,以这个字母开始的串可以删去一些字母从而符合要求,那么此时就要计算…

继续大战dp.2018年11月30日修订,更新一下现在看到这个题目的理解(ps:就现在,poj又503了). 题意分析 这条题目的大意是这样的,问一字符串内最少删去多少的字符使其由给定的若干字符串构成.非常好的一道字符串dp题. 具体的dp解法是什么呢?考虑一下我们删去的这个过程.比如说这个式子 throw thraow 我们要不然删去a,要不然删去 “thraow”才能满足由第一个式子构成的这个条件(空串也算被第一个式子构造了).但是,程序如何知道删去a是使这俩个单词匹配的最优决策? 我们这样…

题目大意:给定一个字符串和一本字典,问至少需要删除多少个字符才能匹配到字典中的单词序列.PS:是单词序列,而不是一个单词 思路:                                                                   动态规划 主要是知道状态方程的含义: 令dp[i]表示从message中第i个字符开始,到第L个字符(结尾处)这段区间所删除的字符数,初始化为dp[L]=0 (dp[i]的下标从0开始) 从message尾部向头部检索匹配,所以是下面的状…

http://poj.org/problem?id=3267 题意 : 给你一个message,是给定字符串,然后再给你字典,让你将message与字典中的单词进行匹配,输出要删掉多少字母. 思路 : 动态规划问题,不仅要找对公式,还要注意一些细节问题 样例解释 : 6是字典里有6个单词,10是给定的字符串的长度,给定的字符串可以与下面的brown和cow匹配,但要删掉两个字母 6 10 browndcodw cow milk white black brown farmer #include<…

山东省ACM多校联盟省赛个人训练第六场 D Rotating Scoreboard https://vjudge.net/problem/POJ-3335 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coac…

带来两题贪心算法的题. 1.给定长度为N的字符串S,要构造一个长度为N的字符串T.起初,T是一个空串,随后反复进行下面两个操作:1.从S的头部删除一个字符,加到T的尾部.2.从S的尾部删除一个字符,加到T的尾部.求你任意采取这两个步骤后能得到的最小字符串T 2.直线上有N个点,点i的位置是xi,这N个点中选择若干个做上标记,对于每个点,在他们距离为R的区域内必须带有标记点,求在满足这个条件的情况下,所需要标记点的最少个数. 1.POJ 3617 Best Cow Line http://poj.…

牛杂技团 题目大意:一群牛想逃跑,他们想通过搭牛梯来通过,现在定义risk(注意可是负的)为当前牛上面的牛的总重量-当前牛的strength,问应该怎么排列才能使risk最小? 说实话这道题我一开始给书上的二分法给弄懵了,后来看了一下题解发现根本不是这么一回事,其实就是个简单的贪心法而已. 这题怎么贪心呢?就是按w+s从小到大排序就好了,证明一下: 1.先证明如果不满足w+s的序列性,无论谁在谁的上面,都会违反题设:(设A在B的上面) 如果 A.s+A.w<B.s+B.w 则如果B.s<m+A…

题目传送门 1 2 题意:有向图,所有点先走到x点,在从x点返回,问其中最大的某点最短路程 分析:对图正反都跑一次最短路,开两个数组记录x到其余点的距离,这样就能求出来的最短路以及回去的最短路. POJ 3268 //#include <bits/stdc++.h> #include <cstdio> #include <queue> #include <algorithm> #include <cstring> using namespace…

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…

树的直径:树上的最长简单路径. 求解的方法是bfs或者dfs.先找任意一点,bfs或者dfs找出离他最远的那个点,那么这个点一定是该树直径的一个端点,记录下该端点,继续bfs或者dfs出来离他最远的一个点,那么这两个点就是他的直径的短点,距离就是路径长度.具体证明见http://www.cnblogs.com/wuyiqi/archive/2012/04/08/2437424.html 其实这个自己画画图也能理解. POJ 1985 题意:直接让求最长路径. 可以用dfs也可以用bfs bfs代…

Description 没有几个人知道,奶牛有她们自己的字典,里面的有W (1 ≤ W ≤ 600)个词,每个词的长度不超过25,且由小写字母组成.她们在交流时,由于各种原因,用词总是不那么准确.比如,贝茜听到有人对她说"browndcodw",确切的意思是"browncow",多出了两个"d",这两个"d"大概是身边的噪音. 奶牛们发觉辨认那些奇怪的信息很费劲,所以她们就想让你帮忙辨认一条收到的消息,即一个只包含小写字母且长…

问题: 链接:http://poj.org/problem?id=3617 思路: 按照字典序比较S和将S反转后的字符串S' 如果S较小,就从S的开头取出一个字符,加到T的末尾(更新下标值) 如果S’较小,从S’的末尾取出一个字符,加到T的末尾(更新下标值) 代码: # include <iostream> # include <string> using namespace std; int N; ]; ]; int main() { cin>>N; ; i <…

Problem 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…