DP的时候记录下是否可以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
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.
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
3 2 1 1 4 1 2 3 4 5 1 1 2 2 2
1 3 2HintFor 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.
#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;
}