标签：des style blog class code c
| 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 }
hoj_10001_朴素DP（LIS）,布布扣,bubuko.com
标签：des style blog class code c
原文地址：http://www.cnblogs.com/sayeter/p/3735561.html
