POJ 2250

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POJ 2250 Compromise(LCS)

POJ 2250 Compromise(LCS)解题报告题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/D 题目: Description In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulf…

poj 2250 Compromise(区间dp)

题目链接:http://poj.org/problem?id=2250 思路分析:最长公共子序列问题的变形,只是把字符变成了字符串,按照最长公共子序列的思路即可以求解. 代码如下: #include <stdio.h> #include <string.h> #define IsEqual(a, b) strcmp((a), (b)) == 0 enum { Left, Up, UpAndLeft }; int XLen, YLen; + ; ], Y[MAX_N][]; int…

POJ 2250 （LCS，经典输出LCS序列 dfs）

题目链接: http://poj.org/problem?id=2250 Compromise Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9284 Accepted: 3972 Special Judge Description In a few months the European Currency Union will become a reality. However, to join the clu…

【POJ 2250】Compromise(最长公共子序列LCS)

题目字符串的LCS,输出解我比较不会,dp的时候记录从哪里转移来的,之后要一步一步转移回去把解存起来然后输出. #include<cstdio> #include<cstring> #define N 102 #define M 32 int n,m,dp[N][N],ans[N][N]; char s1[N][M],s2[N][M],str[N][M]; void solve(){ memset(dp,0,sizeof dp); memset(ans,0,sizeof ans);…

LCS(打印路径) POJ 2250 Compromise

题目传送门题意:求单词的最长公共子序列,并要求打印路径分析:LCS 将单词看成一个点,dp[i][j] = dp[i-1][j-1] + 1 (s1[i] == s2[j]), dp[i][j] = max (dp[i-1][j], dp[i][j-1]) 代码: #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <strin…

POJ 2250(LCS最长公共子序列)

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…

POJ 2250(最长公共子序列变形)

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 not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germa…