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
71 7 3 5 9 4 8
Sample Output
4
题意:输出最长递增子序列的长度
思路：直接裸LIS，第一次使用，使用两种方法
第一种：复杂度n^2
#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int a[1005],dp[1005],n;int LIS(){ int i,j,ans,m; dp[1] = 1; ans = 1; for(i = 2;i<=n;i++) { m = 0; for(j = 1;j<i;j++) { if(dp[j]>m && a[j]<a[i]) m = dp[j]; } dp[i] = m+1; if(dp[i]>ans) ans = dp[i]; } return ans;}int main(){ int i; while(~scanf("%d",&n)) { for(i = 1;i<=n;i++) scanf("%d",&a[i]); printf("%d\n",LIS()); } return 0;}
第二种：nlogn
#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;int a[1005],dp[1005],c[1005],n;int bin(int size,int k){ int l = 1,r = size; while(l<=r) { int mid = (l+r)/2; if(k>c[mid] && k<=c[mid+1]) return mid+1; else if(k<c[mid]) r = mid-1; else l = mid+1; }}int LIS(){ int i,j,ans=1; c[1] = a[1]; dp[1] = 1; for(i = 2; i<=n; i++) { if(a[i]<=c[1]) j = 1; else if(a[i]>c[ans]) j = ++ans; else j = bin(ans,a[i]); c[j] = a[i]; dp[i] = j; } return ans;}int main(){ int i; while(~scanf("%d",&n)) { for(i = 1; i<=n; i++) scanf("%d",&a[i]); printf("%d\n",LIS()); } return 0;}