Longest Ordered Subsequence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 35037 | Accepted: 15372 |
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, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Your program, when given the numeric sequence, must find the length of its longest ordered subsequence.
Input
The first line of input file contains the length of sequence N. The second line contains the elements of sequence - N integers in the range from 0 to 10000 each, separated by spaces. 1 <= N <= 1000
Output
Output file must contain a single integer - the length of the longest ordered subsequence of the given sequence.
Sample Input
7 1 7 3 5 9 4 8
Sample Output
4
大致题意：给出一个数列，求最大不下降子序列的长度。
思路：最优结构：dp[i]以第i个元素为子序列最后一个元素的LIS的长度。
状态转移方程：dp[1]=1;dp[i]=max(dp[j] | j<i && m[j] < m[i] ) +1 ;i>1
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 const int inf = 10001; 5 const int MAX = 1005; 6 int binary_search(int s[],int digit,int length) 7 { 8 int left=0,right=length,mid; 9 while(right!=left) 10 { 11 mid=(left+right)/2; 12 if(digit==s[mid]) 13 return mid; 14 else if(digit<s[mid]) 15 right=mid; 16 else 17 left=mid+1; 18 } 19 return left; 20 } 21 int main() 22 { 23 int n,i,j; 24 while(~scanf("%d",&n)) 25 { 26 int a[MAX]; 27 int s[MAX]; 28 for(i=1;i<=n;i++) 29 scanf("%d",&a[i]); 30 s[0]=-1; 31 int len=1; 32 for(i=1;i<=n;i++) 33 { 34 s[len]=inf; 35 j=binary_search(s,a[i],len); 36 if(j==len) 37 len++; 38 s[j]=a[i]; 39 } 40 printf("%d ",len-1); 41 } 42 return 0; 43 }