uva 10534 Wavio Sequence

Wavio is a sequence of integers. It has some interesting properties.

·  Wavio is of odd length i.e. L = 2*n + 1.

·  The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.

·  The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.

·  No two adjacent integers are same in a Wavio sequence.

For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Wavio sequence which is a subsequence of the given sequence. Consider, the given sequence as :

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1.

Here the longest Wavio sequence is : 1 2 3 4 5 4 3 2 1. So, the output will be 9.

Input

The input file contains less than 75 test cases. The description of each test case is given below: Input is terminated by end of file.

Each set starts with a postive integer, N(1<=N<=10000). In next few lines there will be N integers.

Output

For each set of input print the length of longest wavio sequence in a line.

Sample Input                                   Output for Sample Input

10                                                                                                

1 2 3 4 5 4 3 2 1 10 

19 

1 2 3 2 1 2 3 4 3 2 1 5 4 1 2 3 2 2 1 

5 

1 2 3 4 5

9 
9 
1

题目大意：找出最长的回文子序列。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#define N 10005
using namespace std;
typedef long long ll;
int num[N], dp1[N], dp2[N], S1[N], S2[N];
int main() {
	int n;
	while (scanf("%d", &n) != EOF) {
		for (int i = 0; i < n; i++) {
			scanf("%d", &num[i]);
		}	
		int Max1 = 0, cnt1 = 0;
		for (int i = 0; i < n; i++) {
			if (cnt1 == 0) {
				S1[cnt1++] = num[i];
				dp1[0] = 1;
				continue;
			}
			if (S1[cnt1 - 1] < num[i]) {
				S1[cnt1++] = num[i];
			} else {
				int pos = lower_bound(S1, S1 + cnt1, num[i]) - S1;
				S1[pos] = num[i];
			}
			dp1[i] = cnt1;
		}
		int Max2 = 0, cnt2 = 0;
		for (int i = n - 1; i >= 0; i--) {
			if (cnt2 == 0) {
				S2[cnt2++] = num[i];
				dp2[n - 1] = 1;
				continue;
			}
			if (S2[cnt2 - 1] < num[i]) {
				S2[cnt2++] = num[i];
			} else {
				int pos = lower_bound(S2, S2 + cnt2, num[i]) - S2;
				S2[pos] = num[i];
			}
			dp2[i] = cnt2;
		}	
		int temp = 999999999, ans;
		for (int i = 0; i < n; i++) {
			if (temp > abs(dp1[i] - dp2[i])) {
				temp = abs(dp1[i] - dp2[i]);
				ans = min(dp1[i], dp2[i]);
			}	
		}	
			printf("%d\n", ans * 2 - 1);
	}	
	return 0;
}

uva 10534 Wavio Sequence （最长上升子序列）

原文：http://blog.csdn.net/llx523113241/article/details/44627473