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,…
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…
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: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…
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…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 解题思路:任意先给出两个字符串 abcfbc abfcab,用dp[i][j]来记录当前最长的子序列,则如果有x[i]与y[j]相等的话,则相当于公共子序列的长度在dp[i-1][j-1]上增加1, 如果x[i]与y[j]不相等的话,那么dp[i][j]就取得dp[i][j-1]和dp[i-1][j]中的最大值即可.时间复杂度为O(mn) 反思:大概思路想出来之后,因为dp数组赋初值调了很久,…
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 <…
解题思路:先注意到序列和串的区别,序列不需要连续,而串是需要连续的,先由样例abcfbc         abfcab画一个表格分析,用dp[i][j]储存当比较到s1[i],s2[j]时最长公共子序列的长度 a    b    f    c    a    b 0    0    0    0    0   0    0 a  0    1     1    1    1   1    1 b  0    1     2    2    2   2    2 c  0    1     2  …