HDU5087——Revenge of LIS II(BestCoder Round #16)

【HDU5087——Revenge of LIS II(BestCoder Round #16)】的更多相关文章

HDU5087——Revenge of LIS II(BestCoder Round #16)

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…

hdu5087——Revenge of LIS II

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…

HDU5086——Revenge of Segment Tree(BestCoder Round #16)

Revenge of Segment Tree Problem DescriptionIn computer science, a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point. It is, in principle, a static structure…

hdu 5086 Revenge of Segment Tree(BestCoder Round #16)

Revenge of Segment Tree Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) To…

hdu5087 Revenge of LIS II (dp)

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…

HDU5087 Revenge of LIS II (LIS变形)

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…

HDU 5078 Revenge of LIS II（dp LIS）

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…