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…

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…

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…

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…

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: 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…

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…