HDU 1159 Common Subsequence (动规+最长公共子序列)

【HDU 1159 Common Subsequence (动规+最长公共子序列)】的更多相关文章

HDU 1159 Common Subsequence (动规+最长公共子序列)

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…

HDU 1159 Common Subsequence【dp+最长公共子序列】

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…

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…

Common Subsequence POJ - 1458 最长公共子序列线性DP

#include <iostream> #include <algorithm> #include <string> #include <cstring> #include <cstdio> #define MAX 1005 using namespace std; int ans[MAX][MAX]; int main(){ string s1,s2; while(cin>>s1>>s2) { memset(ans,,s…

HDU 1159 Common Subsequence 最长公共子序列

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…