hdu5087——Revenge of LIS II

Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 444    Accepted Submission(s): 143

Problem Description

In computer science, the longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence's elements are in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence is not necessarily contiguous, or unique.
---Wikipedia

Today, LIS takes revenge on you, again. You mission is not calculating the length of longest increasing subsequence, but the length of the second longest increasing subsequence.
Two subsequence is different if and only they have different length, or have at least one different element index in the same place. And second longest increasing subsequence of sequence S indicates the second largest one while sorting all the increasing subsequences of S by its length.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence.

[Technical Specification]
1. 1 <= T <= 100
2. 2 <= N <= 1000
3. 1 <= Ai <= 1 000 000 000

 

Output

For each test case, output the length of the second longest increasing subsequence.

 

Sample Input

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

 

Sample Output

1
3
2
Hint
For the first sequence, there are two increasing subsequence: [1], [1]. So the length of the second longest increasing subsequence is also 1, same with the length of LIS.
 

 

Source

BestCoder Round #16

 

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这题太坑了，把思路全然引到了求出LIS，然后推断LIS是否唯一上去了
事实上不然。 比方 1 1 2。LIS == 2。可是光这样无法推断出来次大的长度是多少

网上题解是记录每个位置LIS的个数，假设到最后一位LIS仅仅有一个就输出LIS-1，否则去推断LIS上每个位置上LIS是否唯一。不唯一就输出LIS,否则输出LIS - 1

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int dp[1010];
int a[1010];
int num[1010];

int main()
{
	int t, n;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		memset ( dp, 0, sizeof(dp) );
		memset (num, 0, sizeof(num) );
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
			dp[i] = 1;
			num[i] = 1;
		}
		num[n + 1] = 1;
		dp[n + 1] = 1;
		a[n + 1] =  1000000010;
		for (int i = 1; i <= n + 1; i++)
		{
			for (int j = 1; j < i; j++)
			{
				if (a[i] > a[j] && dp[i] < dp[j] + 1)
				{
					num[i] = 1;
					dp[i] = dp[j] + 1;
				}
				else if (a[i] > a[j] && dp[i] == dp[j] + 1)
				{
					num[i]++;
				}
			}
		}
		if (num[n + 1] > 1)
		{
			printf("%d
", dp[n + 1] - 1);
			continue;
		}
		int k = n + 1, i;
		while (k > 0 && num[k] == 1)
		{
			for (i = k - 1; i >= 1; i--)
			{
				if (dp[k] == dp[i] + 1 && a[k] > a[i])
				{
					break;
				}
			}
			k = i;
		}
		if (k == 0)
		{
			printf("%d
", dp[n + 1] - 2);
			continue;
		}
		printf("%d
", dp[n + 1] - 1);
	}
	return 0;
}