Vawio Sequence
1000 ms | 内存限制： 65535
3
Vawio Sequence is very funny，it is a sequence of integers. It has some interesting properties.
· Vawio is of odd length i.e. L = 2*n + 1.
· The first (n+1) integers of Vawio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Vawio sequence makes a strictly decreasing sequence.
· No two adjacent integers are same in a Vawio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Vawio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid Vawio sequence. In this problem, you will be given a sequence of integers. You have to find out the length of the longest Vawio 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.
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
输出
For each set of input print the length of longest Vawio sequence in a line.
样例输入
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
样例输出
9 9 1
//看了大神的才知还可以这样解，我一开始是用LIS写的，但超时。。。
//AC代码
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 10010
using namespace std;
int a[N];
int l[N];
int r[N];
int dp[N];
int main()
{
int n,i,j,k;
while(scanf("%d",&n)!=EOF)
{
memset(dp,0x3f3f3f,sizeof(dp));
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
k=lower_bound(dp,dp+n,a[i])-dp;
dp[k]=a[i];
l[i]=k+1;
}
memset(dp,0x3f3f3f,sizeof(dp));
for(i=n-1;i>=0;i--)
{
k=lower_bound(dp,dp+n,a[i])-dp;
dp[k]=a[i];
r[i]=k+1;
}
int ans=0;
for(i=0;i<n;i++)
ans=max(ans,min(l[i],r[i]));
printf("%d\n",2*ans-1);
}
return 0;
}//我的LIS超时。。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f
#define N 10010
using namespace std;
int a[N];
int b[N];
int c[N];
int main()
{
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
for(i=0;i<n;i++)
{
b[i]=1;
c[i]=1;
}
for(i=n-2;i>=0;i--)
{
for(j=i+1;j<n;j++)
{
if(a[i]<a[j]&&b[i]<b[j]+1)
b[i]=b[j]+1;
if(a[i]>a[j]&&c[i]<c[j]+1)
c[i]=c[j]+1;
}
}
int mm1=0,mm2=0;
for(i=0;i<n;i++)
{
mm1=max(mm1,b[i]);
mm2=max(mm2,c[i]);
}
printf("%d\n",2*min(mm1,mm2)-1);
}
return 0;
}
