hdu 5087 Revenge of LIS II（LIS，思路）

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本文介绍了一种算法解决方案，用于寻找给定序列中第二长的递增子序列。通过对最长递增子序列（LIS）进行分析，确定次长递增子序列的长度，并提供了详细的实现代码。

Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1419 Accepted Submission(s): 486

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

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.

题意：求次长上升子序列

思路：若最长上升子序列唯一，则为LIS，否则为LIS-1

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1010
int a[N];
int dp[N],num[N];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            num[i]=1;
            dp[i]=1;
        }
        int maxn=1;
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<i; j++)
            {
                if(a[i]<a[j]) continue;
                if(a[i]==a[j])
                {
                    dp[i]=max(dp[i],dp[j]);
                    num[i]=max(num[i],num[j]);
                    continue;
                }
                if(dp[i]<dp[j]+1)
                {
                    dp[i]=dp[j]+1;
                    num[i]=num[j];
                }
                else if(dp[i]==dp[j]+1)
                    num[i]+=num[j];
            }
            maxn=max(maxn,dp[i]);
        }
        int s=0;
        for(int i=0; i<n; i++)
            if(dp[i]==maxn)
                s+=num[i];
        if(s==1) printf("%d\n",maxn-1);
        else printf("%d\n",maxn);
    }
    return 0;
}