http://poj.org/problem?id=2533
Longest Ordered Subsequence
Time Limit: 2000MS Memory Limit: 65536K

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
/*

裸的LIS問題： 對于LIS做dp,

下面是根據我的理解寫的求解步驟，不足之處歡迎指出：

1.確定最優狀態：設dp[i]為序列數組中i位置上的最優狀態2. 考慮如何達到最優狀態：即dp[i]是怎么出來的，對于i位置的最優狀態是由i之前的（設為j,0...j < i）最大dp[j]得來的，如果a[i] > a[j],那么dp[i] = max(dp[0...j])+1否則dp[i] = max（dp[0...j]）3.由2推出狀態轉移方程：dp[i] = max（dp[j]+1,dp[i]）(0..j,j<i&&a[i]>a[j])4.考慮邊界：在未dp前，對于每單獨的一個數，dp[i] = 1；5.write code...

*/
AC_code:

#include <iostream> using namespace std; int dp[1005],a[1005];int main(){int n;cin>>n;for(int i = 0; i < n; i++){cin>>a[i];dp[i] = 1;}int ans = 0;for(int i = 0; i <n; i++){for(int j = 0; j < i; j++){if(a[i] > a[j]){dp[i] = max(dp[j]+1,dp[i]);}}ans = max(ans,dp[i]);}cout<<ans<<endl;return 0;}

總結

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