题目传送门 题意:输出两字符串的最长公共子序列长度 分析:LCS(Longest Common Subsequence)裸题.状态转移方程:dp[i+1][j+1] = dp[i][j] + 1; (s[i] == t[i])dp[i+1][j+1] = max (dp[i][j+1], dp[i+1][j]); (s[i] != t[i]) 代码: #include <cstdio> #include <cstring> #include <iostream> #in…

POJ 1458 Common Subsequence(LCS最长公共子序列)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/F 题目: Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43388   Accepted: 17613 Description A subsequen…

POJ 1458 最长公共子序列 题目大意:给出两个字符串,求出这样的一 个最长的公共子序列的长度:子序列 中的每个字符都能在两个原串中找到, 而且每个字符的先后顺序和原串中的 先后顺序一致. Sample Input : abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 分析: 输入两个串s1,s2, 设dp(i,j)表示: s1的左边i个字符形成的子串,与s2左边的j个 字符形成的子串的最长公共子序列的长度(i,j从…

题目传送门 POJ 1458 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 ex…

http://poj.org/problem?id=1458 一道容易的DP,求最长公共子序列的 dp[i][j],代表str1中的第i个字母和str2中的第j个字母的最多的公共字母数 #include <stdio.h> #include <iostream> #include <string.h> using namespace std; ][]={}; int main() { ],str2[]; while(~scanf("%s %s",st…

1.链接地址: http://poj.org/problem?id=1458 http://bailian.openjudge.cn/practice/1458/ 2.题目: Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35411   Accepted: 14080 Description A subsequence of a given sequence is the given…

题目链接:http://poj.org/problem?id=1458 思路分析:经典的最长公共子序列问题(longest-common-subsequence proble),使用动态规划解题. 1)问题定义:给定两个序列X=<X1, X2, ...., Xm>和Y = <Y1, Y2, ...., Yn>,要求求出X和Y长度最长的最长公共子序列: 2)问题分析: <1>动态规划问题都是多阶段决策最优化问题:在这些问题中,问题可以被划分为多个阶段,每个阶段都需要作出一…

http://poj.org/problem?id=1458 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…

POJ1458 Common Subsequence(最长公共子序列LCS) http://poj.org/problem?id=1458 题意: 给你两个字符串, 要你求出两个字符串的最长公共子序列长度. 分析: 本题不用输出子序列,非常easy,直接处理就可以. 首先令dp[i][j]==x表示A串的前i个字符和B串的前j个字符的最长公共子序列长度为x. 初始化: dp全为0. 状态转移: IfA[i]==B[j] then dp[i][j]= dp[i-1][j-1]+1 else dp[…

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…

POJ:http://poj.org/problem?id=1458 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=733 HDU: http://acm.hdu.edu.cn/showproblem.php?pid=1159 题目大意: 给定两串子序列,求最长的公共字串(LCS) 设d( i , j)为A和 B的LCS的长度,则当A[i] = B[j]时, d(i , j)= d(i-1, j-1)+1 ; 否则…

求最长公共子序列LCS,用动态规划求解. UVa的字符串可能含有空格,开始用scanf("%s", s);就WA了一次...那就用gets吧,怪不得要一行放一个字符串呢. (本来想用fgets的,可是又放弃了,形式麻烦.代码长是一小方面,另一方面fgets把'\n'字符也读入,还要做额外的处理...虽然gets有传说中的缓冲区溢出漏洞,不过多加注意一下就好啦,个人认为代码还没大到要用那些工程性的东西的时候) #include <cstdio> #include <cs…

一.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 strict…

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 46387   Accepted: 19045 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

#include <iostream> #include <string> #define MAXN 1000 using namespace std; string s_1; string s_2; void init(); int _m[MAXN][MAXN]; void fun_lcs(); int main() { // freopen("acm.acm","r",stdin); while(cin>>s_1>>…

子序列就是子序列中的元素是母序列的子集,且子序列中元素的相对顺序和母序列相同. 题目要求便是寻找两个字符串的最长公共子序列. dp[i][j]表示字符串s1左i个字符和s2左j个字符的公共子序列的最大长度. 注意s1第i个字符为s1[i-1] 于是有递推公式: 对于abcfbc和abfcab两个字符串,求公共子串的最大长度的过程如图: //#define LOCAL #include <iostream> #include <cstdio> #include <cstring…

LCS #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define clc(a,b) memset(a,b,sizeof(a)) #define LL long long #include<cmath> using namespace std; ][];//表示到i-1,j-1的最长公共长度 int main() { // freopen(&q…

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43132   Accepted: 17472 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

F - F Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u   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 > anothe…

题意:略 求最长公共子串 #include<iostream> #include<cstdio> #include<string> using namespace std; int dp[500][500]; int max(int a,int b) { return a>b?a:b; } int main() { int i,j,a,b; char s1[400],s2[400]; while(scanf("%s%s",s1,s2)!=EOF…

经典的最长公共子序列问题. 状态转移方程为 : if(x[i] == Y[j]) dp[i, j] = dp[i - 1, j - 1] +1 else dp[i, j] = max(dp[i - 1], j, dp[i, j - 1]); 设有字符串X和字符串Y,dp[i, j]表示的是X的前i个字符与Y的前j个字符的最长公共子序列长度. 如果X[i] == Y[j] ,那么这个字符与之前的LCS 一定可以构成一个新的LCS: 如果X[i] != Y[j] ,则分别考察 dp[i  -1][j…

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 65333   Accepted: 27331 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

题目大意:求两个字符串的最长公共子序列 题目思路:dp[i][j] 表示第一个字符串前i位 和 第二个字符串前j位的最长公共子序列 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<iostream> #include<algorithm> #define INF 0x3f3f3f3f #define MAXSIZE 10…

题意:给定两个字符串,让你找出它们之间最长公共子序列(LCS)的长度. 析:很明显是个DP,就是LCS,一点都没变.设两个序列分别为,A1,A2,...和B1,B2..,d(i, j)表示两个字符串LCS长度. 当A[i] = B[j] 时,这个最长度就是上一个长度加1,即:d(i, j) = d(i-1, j-1) + 1; 当A[i] != B[j] 时,那就是前面的最长长度(因为即使后面的不成立,也不会影响前面的),即:d(i, j) = max{d(i-1, j), d(i, j-1)}…

Common Subsequence Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 34477   Accepted: 13631 Description A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..…

最长公共子序列.状态转移方程见代码. #include <iostream> #include <cstdio> #include <cstring> using namespace std; char s1[1005],s2[1005]; int dp[1005][1005]; int main() { while(scanf("%s",s1+1)!=EOF) { scanf("%s",s2+1); memset(dp,0,si…

1808:公共子序列 查看 提交 统计 提问 总时间限制:  1000ms 内存限制:  65536kB 描述 我们称序列Z = < z1, z2, ..., zk >是序列X = < x1, x2, ..., xm >的子序列当且仅当存在 严格上升 的序列< i1, i2, ..., ik >,使得对j = 1, 2, ... ,k, 有xij = zj.比如Z = < a, b, f, c > 是X = < a, b, c, f, b, c >…

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…

最长公共子序列可以用在下面的问题时:给你一个字符串,请问最少还需要添加多少个字符就可以让它编程一个回文串? 解法:ans=strlen(原串)-LCS(原串,反串); Sample Input abcfbc abfcab programming contest abcd mnp Sample Output 4 2 0 代码: #include <stdio.h> #include <string.h> #include <stdlib.h> #include <c…

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…