HDOJ 5087 Revenge of LIS II DP

DP的时候记录下能否够从两个位置转移过来。

。。。

Revenge of LIS II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 393    Accepted Submission(s): 116

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

field=problem&key=BestCoder+Round+%2316&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">BestCoder Round #16

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n;
int a[1100],len[1100];
bool db[1100];

int main()
{
  int T_T;
  scanf("%d",&T_T);
  while(T_T--)
    {
      memset(len,0,sizeof(len));
      memset(db,false,sizeof(db));

      scanf("%d",&n);
      int LIS=-1;
      for(int i=0;i<n;i++)
        {
          scanf("%d",a+i);
          len[i]=1;
          int mxlen=-1,mxp=-1;
          for(int j=0;j<i;j++)
            {
              if(a[j]<a[i])
                {
                  if(len[j]+1>mxlen)
                    {
                      mxlen=len[j]+1; mxp=j;
                    }
                }
            }
          len[i]=max(len[i],mxlen);
          int c1=0;
          for(int j=0;j<i;j++)
            {
              if(a[j]>=a[i]) continue;
              if(len[j]+1==len[i])
                {
                  c1++;
                  if(db[j]==true)
                    {
                      db[i]=true;
                    }
                }
            }
          if(c1>=2)
            {
              db[i]=true;
            }
        }
      for(int i=0;i<n;i++)
        {
          if(LIS<len[i])
            {
              LIS=len[i];
            }
        }
      bool flag=false;
      int c1=0;
      for(int i=0;i<n&&flag==false;i++)
        {
          if(LIS==len[i])
            {
              if(db[i]==true) flag=true;
              c1++;
            }
        }
      if(c1>=2||flag)
        {
          printf("%d\n",LIS);
        }
      else printf("%d\n",max(1,LIS-1));
    }
  return 0;
}