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UVA 10534 Wavio Sequence(LIS)
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
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Problemsetter: Md. Kamruzzaman, Member of Elite Problemsetters‘ Panel
对每个点前后LIS:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<limits.h> typedef long long LL; using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) const int maxn=10000+100; int a[maxn],s[maxn]; int t1[maxn],t2[maxn]; int n; int main() { std::ios::sync_with_stdio(false); while(~scanf("%d",&n)) { REPF(i,1,n) scanf("%d",&a[i]); REPF(i,1,n) { s[i]=INT_MAX; int tt=lower_bound(s+1,s+1+i,a[i])-s; t1[i]=tt; s[tt]=a[i]; } for(int i=n;i>=1;i--) { s[n-i+1]=INT_MAX; int tt=lower_bound(s+1,s+n-i+2,a[i])-s; t2[i]=tt; s[tt]=a[i]; } int ans=1; REPF(i,1,n) { int temp=min(t1[i],t2[i])*2-1;//合并的时候一定要注意前后长度一样所以取min(t1[i],t2[i]) if(temp>ans) ans=temp; } printf("%d\n",ans); } return 0; }
UVA 10534 Wavio Sequence(LIS)