求一个数列的最长上升序列 动态规划法:O(n^2) //DP int LIS(int a[], int n) { int DP[n]; int Cnt=-1; memset(DP, 0, sizeof(DP)); for(int i=0; i<n; i++ ) { for(int j=0; j<i; j++ ) { if( a[i]>a[j] ) { DP[i] = max(DP[i], DP[j]+1); Cnt = max(DP[i], Cnt);//记录最长序列所含元素的个数 }…

#include<iostream> using namespace std ; ; int f[N]; int a[N]; int n; int main() { cin>>n; ; i<=n; i++) cin>>a[i]; ; i<=n; i++) { f[i]=; ; j<=i; j++) { if(a[j]<a[i]) { f[i]=max(f[i],f[j]+); } } } ; ; i<=n; i++) res=max(res…

Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 47465   Accepted: 21120 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ...…

D - POJ 2533 经典DP-最长上升子序列 A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. Fo…

Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N.…

Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence ( a1, a2, ..., aN) be any sequence ( ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, seq…

  Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, seq…

作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4098562.html 题目链接:poj 2533 Longest Ordered Subsequence 最长递增子序列 使用$len[i]$表示序列中所有长度为$i$的递增子序列中最小的第$i$个数的值为$len[i]$.对于序列的第j个数$arr[j]$,在$len$中二分查找,找到最后一个小于$arr[j]$的数$len[k]$,如果$len[k]$是序列$len$中最后的一个数,那…

POJ 3903    Stock Exchange  (E - LIS 最长上升子序列) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/E 题目: Description The world financial crisis is quite a subject. Some people are more relaxed while others are quite anxious. John…

POJ 1887Testingthe CATCHER (LIS:最长下降子序列) http://poj.org/problem?id=3903 题意: 给你一个长度为n (n<=200000) 的数字序列, 要你求该序列中的最长(严格)下降子序列的长度. 分析:        读取全部输入, 将原始数组逆向, 然后求最长严格上升子序列就可以. 因为n的规模达到20W, 所以仅仅能用O(nlogn)的算法求.        令g[i]==x表示当前遍历到的长度为i的全部最长上升子序列中的最小序列末…