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 length9. But1, 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 be9.
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 beN 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 |
代码:
#include <map> #include <set> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <iomanip> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> #define ll long long #define mod 1000000007 #define mem(a) memset(a,0,sizeof(a)) using namespace std; const int maxn = 10000 + 5 , inf = 0x3f3f3f3f; int a[maxn],up[maxn],down[maxn],dp[maxn]; int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int n; while(scanf("%d",&n)!=EOF){ int ans = 1 ; mem(a); for(int i = 0 ; i < n ; i++ ) scanf("%d",&a[i]); int cnt = 0; dp[0] = a[0]; up[0] = 1; for(int i = 1 ; i < n ; i++){ if(a[i]>dp[cnt]) dp[++cnt] = a[i]; if(a[i]<dp[cnt]){ int k = lower_bound(dp,dp+cnt+1,a[i]) - dp ; dp[k] = a[i]; } up[i]=cnt+1; } cnt = 0 ; mem(dp); dp[0] = a[n-1]; down[n-1] = 1; for(int i = n - 2 ; i>= 0 ; i--){ if(a[i]>dp[cnt]) dp[++cnt] = a[i]; if(a[i]<dp[cnt]){ int k = lower_bound(dp,dp+cnt+1,a[i]) - dp ; dp[k] = a[i]; } down[i]=cnt+1; } for(int i = 0 ; i<n ; i++){ if(up[i]==down[i]&&up[i]>ans) ans = up[i]; } printf("%d\n",2*(ans)-1); } }