首页 > 代码库 > hoj_10001_朴素DP(LIS)
hoj_10001_朴素DP(LIS)
Longest Ordered Subsequence |
Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB |
Total submit users: 1937, Accepted users: 1621 |
Problem 10001 : No special judgement |
Problem 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. |
Input |
The first line of input 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 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 |
这是求lis的一道题,也一个最基本的O(n*n)的一维DP,更是我算法之路的开始~~~~~~
对于给定的a[N]数组,从1到n每次分别求对应a[i]的lis,最后即为最终的lis。
AC代码如下:
1 #include<cstdio> 2 using namespace std; 3 int main() 4 { 5 int t; 6 int n; 7 int a[1005],b[1005]; 8 while(scanf("%d",&n)!=EOF) 9 { 10 for(int i=0;i<n;i++) 11 scanf("%d",&a[i]); 12 b[0]=1; 13 for(int i=1;i<n;i++) 14 { 15 b[i]=1; 16 for(int j=0;j<i;j++) 17 { 18 if(a[i]>a[j] && b[j]+1>b[i]) 19 { 20 b[i]=b[j]+1; 21 } 22 } 23 } 24 t=b[0]; 25 for(int i=0;i<n;i++) 26 { 27 if(b[i]>t) 28 t=b[i]; 29 } 30 printf("%d\n",t); 31 } 32 return 0; 33 }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。