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: 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…

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…

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…

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…

Best Cow Fences Time Limit: 1000MS Memory Limit: 30000K Description Farmer John's farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000. FJ wants to build a fence around a…

bitset的经典优化,即把可行性01数组的转移代价降低 bitset的适用情况,当内层状态只和外层状态的上一个状态相关,并且内层状态的相关距离是一个固定的数,可用bitset,换言之,能用滚动数组是能用bitset优化的前提 /* dp[i,j][0|1|2]表示p串的第i位,s串的第j位相匹配,pi和pi-1换,pi不换,pi和pi+1换的状态下是否能匹配 dp[i,j][0] = dp[i-1,j-1][2] & p[i-1]==s[j] dp[i,j][1] = (dp[i-1,j-1]…

在输入的单词中删除或替换或插入一个字符,看是否在字典中.直接暴力,172ms.. #include <stdio.h> #include <string.h> ]; ][], s[]; bool del(char s1[], char s2[]) { ; , j = ; s1[i] && s2[j]; i++, j++) { if(s1[i] != s2[j]) { if(isdel) ; j--; isdel = ; } } ; } bool rep(char s…