本文主要是介绍NYOJ 763 Vawio Sequence，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
OJ题目 ： 戳这里~
描述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.
- 输入
- 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
#define inf 100000000
int main()
{int dp[10002];int inc[10002];int dec[10002];int x[10002];int n ;while(scanf("%d",&n) != EOF){fill(dp, dp + n, inf);for(int i = 0;i < n;i++)scanf("%d" , &x[i]);for(int i = 0;i < n;i++){int id = lower_bound(dp, dp + n, x[i]) - dp;dp[id] = x[i];inc[i] = id + 1;}fill(dp, dp + n, inf);for(int i = n - 1;i >= 0;i--){int id = lower_bound(dp, dp + n, x[i]) - dp;dp[id] = x[i];dec[i] = id + 1;}int ans = 0;for(int i = 0;i < n;i++){ans = Max(ans , ((Min(inc[i] , dec[i]) << 1) - 1));}printf("%d\n",ans);}return 0;
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