POJ 2533: Longest Ordered Subsequence

【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 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
【题意】
**求最长上升子序列
以dp[i]表示以a[i]为序列结尾的最长上升子序列长度
则 递推方程：dp[i] = max(dp[k]+1) for everya[k]

#include <cstdio>const int MAX = 1000 + 10;int a[MAX], dp[MAX];int main() {    int n;    while (scanf("%d", &n) != EOF) {        for (int i = 0; i < n; ++i) {            scanf("%d", &a[i]);        }        int ans = 1;        for (int i = 0; i < n; ++i) {            // 置初值            dp[i] = 1;            // 遍历所有满足 j < i && a[j] < a[i] 之条件者，            // 并取其最大值            for (int j = i - 1; j >= 0; --j) {                if (a[i] > a[j] && dp[j] + 1 > dp[i]) {                //如果此条件成立，就将位置为 j 的数加入最长子序列                    dp[i] = dp[j] + 1;                }            }            // 同步记录最长序列长度            if (dp[i] > ans) {                ans = dp[i];            }        }        printf("%d\n", ans);    }    return 0;}