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…

题目链接:http://poj.org/problem?id=2533 Time Limit: 2000MS Memory Limit: 65536K 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, ...,…

两种算法 1.  O(n^2) #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]; ]; int main() { int n, maxn; while(scanf("%d", &n) != EOF) { maxn = ; ; i < n; i++) { scanf("%d", &a[i]); dp[i]…

传送门: http://poj.org/problem?id=2533 Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 61731   Accepted: 27632 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the…

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…

传送门 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, s…

Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 38980   Accepted: 17119 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ...…

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, ...…

一.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…

Language: Default Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 33986   Accepted: 14892 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric seq…

Description A numeric sequence of ai is ordered ifa1 <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, sequence (1…

Given an unsorted array of integers, find the number of longest increasing subsequence. Example 1: Input: [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7]. Example 2: Input: [2,2,2,2,2] Outpu…

Given an unsorted array of integers, find the number of longest increasing subsequence. Example 1: Input: [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7]. Example 2: Input: [2,2,2,2,2] Outpu…

题目链接:http://poj.org/problem?id=2533 思路分析:该问题为经典的最长递增子序列问题,使用动态规划就可以解决: 1)状态定义:假设序列为A[0, 1, .., n],则定义状态dp[i]为以在所有的递增子序列中以A[i]为递增子序列的最后一个数字的所有递增子序列中的最大长度: 如:根据题目,在所有的以3结尾的递增子序列有[3]和[1, 3],所以dp[2] =2; 2)状态转移方程:因为当A[j] < A[i]时(0<= j < i),dp[i] = Max…

题目描述:LIS(Longest Increasing Subsequence)模板题 分析:O(n^2)的方法 状态表示:d[i]表示以i结尾的最长上升子序列长度 转移方程:d[i]=max{ 1,d(j)+1 } ( j=1,2,3,...,i-1且A[j]<A[i] ) 即A[j]<A[i],d[i]=d[j]+1 A[j]>=A[i],d[i]=1 #include<cstdio> int main() { ],d[]; scanf("%d",&a…

题目传送门 题意:LIS(Longest Increasing Subsequence)裸题 分析:状态转移方程:dp[i] = max (dp[j]) + 1   (a[j] < a[i],1 <= j < i) 附带有print输出路径函数 代码: #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N = 1e4 + 10…

题目大意:求一个数列的最长上升子序列(严格上升). 解题思路: 方法一:O(n^2) dp[i]:表示处理到第i个位置,序列的最长上升子序列末尾为i的长度: a[]数组存储原序列 dp[i] = max{dp[j]+1},a[i]>a[j],0≤j≤i 方法二:O(nlogn) 复杂度降低其实是因为这个算法里面用到了二分搜索.本来有N个数要处理是O(n),每次计算要查找N次还是O(n),一共就是O(n^2):现在搜索换成了O(logn)的二分搜索,总的复杂度就变为O(nlogn)了. 这个算法的…

1.链接地址: http://poj.org/problem?id=2533 http://bailian.openjudge.cn/practice/2757 2.题目: 总Time Limit: 2000ms Memory Limit: 65536kB Description 一个数的序列bi,当b1 < b2 < ... < bS的时候,我们称这个序列是上升的.对于给定的一个序列(a1, a2, ..., aN),我们可以得到一些上升的子序列(ai1, ai2, ..., aiK)…

d.最长上升子序列 s.注意是严格递增 c.O(nlogn) #include<iostream> #include<stdio.h> using namespace std; ; int a[MAXN],b[MAXN]; //b[k]是序列a中所有长度为k的递增子序列中的最小结尾元素值 //用二分查找的方法找到一个位置,使得num>b[i-1]并且num<b[i],并用num代替b[i] int Search(int num,int low,int high){ in…

题目:http://poj.org/problem?id=2533 题意:最长上升子序列.... 以前做过,课本上的思想 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { ]; ],i,j,n; int nmax; cin>>n; ; i<=n; i++) { cin>&…

链接:http://poj.org/problem?id=2533 题解 #include<iostream> using namespace std; ]; //存放数列 ]; //b[i]表示以a[i]为结尾的子序列的最大长度 int main(){ int n; scanf("%d",&n); ;i<n;i++) scanf("%d",&a[i]); dp[]=; ;i<n;i++){ ;j<i;j++) //对于…

题目链接 最长上升子序列O(n*log(n))的做法,只能用于求长度不能求序列. #include <iostream> #define SIZE 1001 using namespace std; int main() { int i, j, n, top, temp; int stack[SIZE]; while(cin >> n) { top = ; /* 第一个元素可能为0 */ stack[] = -; ; i < n; i++) { cin >> te…

最长公共自序列LIS 三种模板,但是邝斌写的好像这题过不了 N*N #include <iostream> #include <cstdio> #include <cstring> using namespace std; ; ],dp[],n; int Lis(){ dp[]=; ; ; ;i<=n;i++){ temp=; ;j<i;j++){ if(dp[j]>temp&&a[i]>a[j]){ temp=dp[j]; }…

题目链接 最长上升子序列O(n*log(n))的做法,只能用于求长度不能求序列. #include <iostream> #include <algorithm> using namespace std; ; int s[N],x; int main() { int n; while(cin>>n){ ; ;i<n;i++){ cin>>x; ||s[top-]<x) s[top++]=x; else s[upper_bound(s,s+top,…

Longest Increasing Subsequence 最长递增子序列 子序列不是数组中连续的数. dp表达的意思是以i结尾的最长子序列,而不是前i个数字的最长子序列. 初始化是dp所有的都为1,最终的结果是求dp所有的数值的最大值. class Solution { public: int lengthOfLIS(vector<int>& nums) { int length = nums.size(); ) ; vector<); int max_num; ;i <…

Longest Ordered Subsequence Time Limit: 2 Seconds      Memory Limit: 65536 KB 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), whe…

Given an unsorted array of integers, find the length of longest increasing subsequence. For example, Given [10, 9, 2, 5, 3, 7, 101, 18], The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than…

Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return the length of the LIS. Have you met this question in a real interview?     Example For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3 For [4, 2, 4,…