UVa 10534 - Wavio Sequence LIS
2013-04-24 21:15
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Problem D
Wavio Sequence
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
Wavio is a sequence of integers. It has some interesting properties.
· Wavio is of odd length i.e. L = 2*n + 1.
· The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
· No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to
find out the length of the longest Wavio 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.
Input
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.
Output
For each set of input print the length of longest wavio sequence in a line.
--------------------------------------------
f(i)表示从1到i的最长递增子序列,g(i)表示从n到i的最长递增子序列。
ans=min( min(g(i),g(i))*2-1 ) 1<=i<=n
求LIS复杂度为O(nlogn) 枚举i复杂度为O(n) 总时间复杂度O(nlogn)
-------------------------------------------
Wavio Sequence
Input: Standard Input
Output: Standard Output
Time Limit: 2 Seconds
Wavio is a sequence of integers. It has some interesting properties.
· Wavio is of odd length i.e. L = 2*n + 1.
· The first (n+1) integers of Wavio sequence makes a strictly increasing sequence.
· The last (n+1) integers of Wavio sequence makes a strictly decreasing sequence.
· No two adjacent integers are same in a Wavio sequence.
For example 1, 2, 3, 4, 5, 4, 3, 2, 0 is an Wavio sequence of length 9. But 1, 2, 3, 4, 5, 4, 3, 2, 2 is not a valid wavio sequence. In this problem, you will be given a sequence of integers. You have to
find out the length of the longest Wavio 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.
Input
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.
Output
For each set of input print the length of longest wavio sequence in a line.
Sample Input Output for Sample Input
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 | 9 9 1 |
f(i)表示从1到i的最长递增子序列,g(i)表示从n到i的最长递增子序列。
ans=min( min(g(i),g(i))*2-1 ) 1<=i<=n
求LIS复杂度为O(nlogn) 枚举i复杂度为O(n) 总时间复杂度O(nlogn)
-------------------------------------------
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int OO=1e9; int a[11111]; int b[11111]; int f[11111]; int g[11111]; int d[11111]; int n; int main() { while (~scanf("%d",&n)) { memset(f,0,sizeof(f)); memset(g,0,sizeof(g)); memset(d,0,sizeof(d)); for (int i=1;i<=n;i++) { scanf("%d",&a[i]); b[n-i+1]=a[i]; } for (int i=1;i<=n;i++) d[i]=OO; for (int i=1;i<=n;i++) { int k=lower_bound(d+1,d+n+1,a[i])-d; f[i]=k; d[k]=a[i]; } for (int i=1;i<=n;i++) d[i]=OO; for (int i=1;i<=n;i++) { int k=lower_bound(d+1,d+n+1,b[i])-d; g[n-i+1]=k; d[k]=b[i]; } //for (int i=1;i<=n;i++) cerr<<f[i]<<" ";cerr<<endl; //for (int i=1;i<=n;i++) cerr<<g[i]<<" ";cerr<<endl; int ans=0; for (int i=1;i<=n;i++) { int tmp=min( f[i], g[i] ); tmp=(tmp-1)*2+1; if (tmp>ans) ans=tmp; } printf("%d\n",ans); } return 0; }
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