这题确实很棒..又是无想法..其实是AC自动机+DP的感觉,但是只有一个串,用kmp就行了. dp[i][j][k],k代表前缀为virus[k]的状态,len表示其他所有状态串,处理出Ac[len][26]数组来,DP就可以了.状态转移那里一直没想清楚,wa了很多次,记录路径倒是不复杂,瞎搞搞就行. #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include&…

Common Subsequence Time Limit: 2 Sec  Memory Limit: 64 MBSubmit: 951  Solved: 374 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = <X1, x2, ..., xm>another sequence…

Problem C: Longest Common Subsequence Sequence 1: Sequence 2: Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences: abcdgh…

题目链接:http://poj.org/problem?id=1458 Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 55099   Accepted: 22973 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left ou…

题意:给定两个串,求出两个串的最长公共子序列,要求该公共子序列不包含virus串. 用dp+kmp实现 dp[i][j][k]表示以i结尾的字符串和以j结尾的字符串的公共子序列的长度(其中k表示该公共子序列的与virus的匹配程度)很显然,当k==strlen(virus)时,该公共子序列不是我们所求得 当添加一个字符时,如果失配,这时不能让k直接等于0,而是要用kmp给k一个合理的值. #include <stdio.h> #include <string.h> + ; int…

Common Subsequence A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a s…

Given two strings, find the longest common subsequence (LCS). Your code should return the length of LCS. Example For "ABCD" and "EDCA", the LCS is "A" (or "D", "C"), return 1. For "ABCD" and &quo…

题目链接 基础的最长公共子序列 #include <bits/stdc++.h> using namespace std; ; char c[maxn],d[maxn]; int dp[maxn][maxn]; int main() { while(scanf("%s%s",c,d)!=EOF) { memset(dp,,sizeof(dp)); int n=strlen(c); int m=strlen(d); ;i<n;i++) ;j<m;j++) if(c…

http://poj.org/problem?id=1458 用dp[i][j]表示处理到第1个字符的第i个,第二个字符的第j个时的最长LCS. 1.如果str[i] == sub[j],那么LCS长度就可以+1,是从dp[i - 1][j - 1] + 1,因为是同时捂住这两个相同的字符,看看前面的有多少匹配,+1后就是最大长度. 2.如果不同,那怎么办? 长度是肯定不能增加的了. 可以考虑下删除str[i] 就是dp[i - 1][j]是多少,因为可能i - 1匹配了第j个.也可能删除sub…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39559    Accepted Submission(s): 18178 Problem Description A subsequence of a given sequence is the given sequence with some el…