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Longest Ordered Subsequence
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 34454 | Accepted: 15135 |
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).
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
最长上升子序列。。orz 傻逼竟然直接把dp[n]输出了 后来wa了一时还没反应过来。。
dp[i]代表以i为结尾的最长上升子序列的长度，but dp[n]不一定最长。。事实上整个dp数组就是无序的了。。
能够sort后输出
O（n*n）渣比写法
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <string> #include <cctype> #include <vector> #include <cstdio> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #define ll long long #define maxn 1010 #define pp pair<int,int> #define INF 0x3f3f3f3f #define max(x,y) ( ((x) > (y)) ?(x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;int n,dp[maxn],a[maxn];void solve(){ for(int i=2;i<=n;i++) for(int j=1;j<i;j++) if(a[i]>a[j]&&dp[i]<=dp[j]) dp[i]=dp[j]+1; sort(dp+1,dp+n+1); printf("%d\n",dp[n]);}int main(){ while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) { dp[i]=1; scanf("%d",&a[i]); } solve(); } return 0;}
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