POJ 2533 Longest Ordered Subsequence 简单DP
题目链接：http://poj.org/problem?id=2533
题目大意：求最长递增子序列。。。
解题思路：dp[i] = dp[j] + 1 a[i] > a[j] && dp[j] + 1 > dp[i]
注意初始化为1！
代码：
1 const int inf = 0x3f3f3f3f;
2 const int maxn = 1e3 + 5;
3 int n, a[maxn];
4 int dp[maxn];
5
6 void solve(){
7 memset(dp, 0, sizeof(dp));
8 for(int i = 0; i <= n; i++) dp[i] = 1;
9 int ans = 1;
10 for(int i = 1; i <= n; i++){
11 for(int j = 1; j < i; j++){
12 if(a[j] < a[i] && dp[j] + 1 > dp[i])
13 dp[i] = dp[j] + 1;
14 }
15 ans = max(ans, dp[i]);
16 }
17 printf("%d\n", ans);
18 }
19 int main(){
20 scanf("%d", &n);
21 for(int i = 1; i <= n; i++){
22 scanf("%d", &a[i]);
23 }
24 solve();
25 }
题目：
Longest Ordered Subsequence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 54539 | Accepted: 24426 |
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.
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
Northeastern Europe 2002, Far-Eastern Subregion
