Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18387    Accepted Submission(s): 7769 Problem Description A subsequence of a given sequence is the given sequence with some el…

题意:给定两行字符串,求最长公共子序列. 析:dp[i][j] 表示第一串以 i 个结尾和第二个串以 j 个结尾,最长公共子序列,剩下的就简单了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <…

HDU 1159 题目大意:给定两个字符串,求他们的最长公共子序列的长度 解题思路:设字符串 a = "a0,a1,a2,a3...am-1"(长度为m), b = "b0, b1, b2, b3 ... bn-1"(长度为n), 它们的最长公共子序列为c = "c0, c1, c2, ... ck-1",长度为k, dp[i][j]定义为子串 "a0,a1,...,ai-1" 和 子串"b0,b1,...,bj-1…

HDU 1159 Common Subsequence 最长公共子序列 题意 给你两个字符串,求出这两个字符串的最长公共子序列,这里的子序列不一定是连续的,只要满足前后关系就可以. 解题思路 这个当然要使用动态规划了. 这里\(dp[i][j]\)代表第一个串的前\(i\)个字符和第二个串的前\(j\)个字符中最长的公共子序列的最长长度,递推关系如下: \[ d[i][j]= \begin{cases} dp[i-1][j-1]+1 & \text{if} &str1[i]==str2[j…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47676    Accepted Submission(s): 21890 Problem Description A subsequence of…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37551    Accepted Submission(s): 17206 Problem Description A subsequence of…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18201    Accepted Submission(s): 7697 Problem Description A subsequence of…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25416    Accepted Submission(s): 11276 Problem Description A subsequence of…

链接: http://acm.hdu.edu.cn/showproblem.php?pid=1159 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28195#problem/A Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17621…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 Problem Description 给定序列的子序列是给定的序列,其中有一些元素(可能没有)被遗漏. 给定一个序列X = <x1,x2,...,xm>如果存在严格递增的序列<i1,i2,...,则另一个序列Z = <z1,z2,...,zk>是X的子序列. ...,ik>的索引,使得对于所有j = 1,2,...,k,xij = zj. 例如,Z = <a,…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37725    Accepted Submission(s): 17301 Problem Description A subsequence of a given sequence is the given sequence with some el…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24489    Accepted Submission(s): 10823 Problem Description A subsequence of a given sequence is the given sequence with some e…

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…

2017-08-06 15:41:04 writer:pprp 刚开始学dp,集训的讲的很难,但是还是得自己看,从简单到难,慢慢来(如果哪里有错误欢迎各位大佬指正) 题意如下: 给两个字符串,找到其中大的公共子序列,每个样例输出一个数: 最长公共子串(Longest Common Substirng)和最长公共子序列(Longest Common Subsequence,LCS)的区别为: 子串是串的一个连续的部分,子序列则是从不改变序列的顺序,而从序列中去掉任意的元素而获得新的序列: 也就是说…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34819 Accepted Submission(s): 15901 Problem Description A subsequence of a given sequence is the given sequence with some element…

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 strictly increasing sequ…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18765    Accepted Submission(s): 7946 Problem Description A subsequence of a given sequence is the given sequence with some ele…

Problem 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 Z = <z1, z2, ..., zk> is a subsequence of X if there exists a stri…

题目链接 Problem 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 Z = <z1, z2, ..., zk> is a subsequence of X if there exists a…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22698    Accepted Submission(s): 9967 Problem Description A subsequence of a given sequence is the given sequence with some el…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55301    Accepted Submission(s): 25537 Problem Description A subsequence of a given sequence is the given sequence with some el…

Problem 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 Z = <z1, z2, -, zk> is a subsequence of X if there exists a strictly…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 题意: 求最长公共子序列. 题解: (LCS模板题) 表示状态: dp[i][j] = max len of LCS a串匹配到第i位,b串匹配到第j位,此时的最长公共子序列长度. 如何转移: 首先,一个明显的决策是,如果a[i] == b[j],那么此一定要匹配.(贪心) 所以分两种情况: (1)a[i] == b[j]:dp[i][j] = dp[i-1][j-1] + 1 (2)a[i]…

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…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 题解 #include<iostream> #include<cstring> using namespace std; ],B[]; //存放字符串 ][]; //存放到字符串a第 i+1个字符,字符串b第 j+1个字符为止的最大长度公共子序列的长度 int main(){ while(~scanf("%s%s",A,B)){ int alen=strlen(A…

题目链接:https://vjudge.net/contest/124428#problem/A 题目大意:给出两个字符串,求其最长公共子序列的长度. 最长公共子序列算法详解:https://blog.csdn.net/hrn1216/article/details/51534607     (其中的图解很详细)   根据图解理解下面代码 #include<cstdio> #include <string> #include<cstring> #include<i…

题意: 两个字符串,判断最长公共子序列的长度. 思路: 直接看代码,,注意边界处理 代码: char s1[505], s2[505]; int dp[505][505]; int main(){ while(scanf("%s%s",s1,s2)!=EOF){ int l1=strlen(s1); int l2=strlen(s2); mem(dp,0); dp[0][0]=((s1[0]==s2[0])?1:0); rep(i,1,l1-1) if(s1[i]==s2[0]) dp…

1. 问题描述 子串应该比较好理解,至于什么是子序列,这里给出一个例子:有两个母串 cnblogs belong 比如序列bo, bg, lg在母串cnblogs与belong中都出现过并且出现顺序与母串保持一致,我们将其称为公共子序列.最长公共子序列(Longest Common Subsequence, LCS),顾名思义,是指在所有的子序列中最长的那一个.子串是要求更严格的一种子序列,要求在母串中连续地出现.在上述例子的中,最长公共子序列为blog(cnblogs, belong),最长公…

HDOJ 1159 Common Subsequence[DP] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44280 Accepted Submission(s): 20431 Problem Description A subsequence of a given sequence is the given sequence wit…

Common Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39661    Accepted Submission(s): 18228 Problem Description A subsequence of a given sequence is the given sequence with some el…