标签：poj 2533 ++ ota esc others plm enc ace mis
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 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 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
Source
分析：
完全裸露的LIS
code：
#include<stdio.h>
#define MAXN 1000
int dp[MAXN+10],a[MAXN+10];//a数组记录输入的序列
int main()
{
int n,i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
dp[1]=1;
for(i=2;i<=n;i++)
{
int temp=0;
for(j=1;j<i;j++)
if(a[i]>a[j])
if(temp<dp[j])
temp=dp[j];
dp[i]=temp+1;
}
int maxlen=0;
for(i=1;i<=n;i++)
if(maxlen<dp[i])
maxlen=dp[i];
printf("%d\n",maxlen);
return 0;
}
POJ 2533 Longest Ordered Subsequence（裸LIS）
标签：poj 2533 ++ ota esc others plm enc ace mis
原文地址：https://www.cnblogs.com/yinbiao/p/9347993.html
