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]…

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, ...,…

  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: 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…

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…

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…

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…

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…

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 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.…

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…

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…

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: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>&…

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)…

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…

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…

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…

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 Common Subsequence最长公共子序列: 每个dp位置表示的是第i.j个字母的最长公共子序列 class Solution { public: int findLength(vector<int>& A, vector<int>& B) { int len1 = A.size(); int len2 = B.size(); || len2 == ) ; vector<vector<,vector<)); ;i <=…

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 an unsorted array of integers, find the length of longest increasing subsequence. Example: Input: [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. Note: There may be more…

两个都是最长上升子序列,所以就放一起了 1631 因为长度为40000,所以要用O(nlogn)的算法,其实就是另用一个数组c来存储当前最长子序列每一位的最小值,然后二分查找当前值在其中的位置:如果当前点不能作为当前最长子序列的最大值,则更新找到值为两者间的较小值. 2533 就是一个裸的最长上升子序列...这里就不多说了,直接dp就好... 1611: #include <iostream> #include <cstring> #include <cstdio> #…