题目链接:pid=5087">http://acm.hdu.edu.cn/showproblem.php?pid=5087 题意: 求第二长的最长递增序列的长度 分析: 用step[i]表示以i结尾的最长上升序列的长度,dp[i]表示到i的不同的最长的子序列的个数 然后最后推断最长的子序列的个数是否大于1是的话输出Max,否则输出Max-1 代码例如以下: #include<cstdio> #include<cstring> #include<algorith…

链接:hdu 5087 题意:求第二大的最长升序子序列 分析:这里的第二大指的是,全部的递增子序列的长度(包含相等的), 从大到小排序后.排在第二的长度 cid=546" style="color:rgb(106,57,6); text-decoration:none">BestCoder Round #16 上的第二题,注意  1 1 2 这组数据,答案应为2 思路1.每次将最长的两个上升子序列长度记录.最后再排序,取第二大的就可以 思路2.假设最长的上升子序列长度(…

Revenge of LIS II Problem DescriptionIn computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence…

Revenge of LIS II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 444    Accepted Submission(s): 143 Problem Description In computer science, the longest increasing subsequence problem is to fi…

DP的时候记录下能否够从两个位置转移过来. ... Revenge of LIS II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 393    Accepted Submission(s): 116 Problem Description In computer science, the longest increasing su…

称号:hdoj 5087 Revenge of LIS II 题意:非常easy,给你一个序列,让你求第二长单调递增子序列. 分析:事实上非常easy.不知道比赛的时候为什么那么多了判掉了. 我们用O(n^2)的时间求单调递增子序列的时候,里面在加一层循环维护sum数组.表示前面有几个能够转移当当前,求前面sum的和保存到当前. 最后求最后一个sum[n-1]是否为1就ok.为1的话在最长的基础上减一,否则就是最长的. AC代码: #include <iostream> #include &l…

Revenge of Nim II Problem DescriptionNim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided th…

位运算.. .. Revenge of Nim II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 229    Accepted Submission(s): 79 Problem Description Nim is a mathematical game of strategy in which two players take…

只要理解了LIS,这道题稍微搞一下就行了. 求LIS(最长上升子序列)有两种方法: 1.O(n^2)的算法:设dp[i]为以a[i]结尾的最长上升子序列的长度.dp[i]最少也得是1,就初始化为1,则dp[i]=max(dp[i],dp[j]+1)(其中j<i且a[j]<a[i]). int gao() { ; ;i<n;i++) { dp[i]=; ;j<i;j++) { if(a[j]<a[i]) { dp[i]=max(dp[i],dp[j]+); } } ans=ma…

Problem Description In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as po…