Common Subsequence POJ-1458 //最长公共子序列问题 #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<string> using namespace std; string a,b; int dp[1003][1003]; int main(){ while(cin>>a>>b){…
compromise Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Description In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is…
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…
LCS最长公共子序列 模板代码: #include <iostream> #include <string.h> #include <string> using namespace std; int dp[110][110]; int main() { string a,b; memset(dp,0,sizeof(dp)); cin>>a>>b; int lena = a.size(); int lenb = b.size(); for(int…
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…
区别最长公共子串(连续) ''' LCS 最长公共子序列 ''' def LCS_len(x, y): m = len(x) n = len(y) dp = [[0] * (n + 1) for i in range(m + 1)] B = [[' '] * (n + 1) for i in range(m + 1)] for i in range(1, m + 1): for j in range(1, n + 1): if x[i - 1] == y[j - 1]: dp[i][j] = d…
