POJ 1458 1159

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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…

LCS POJ 1458 Common Subsequence

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）

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(dp)

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…

POJ 1458 Common Subsequence （动态规划）

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…

HDU 1159 && POJ 1458

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…

HDU 1159 Common Subsequence(POJ 1458)

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…