UVA 10534 - Wavio Sequence(DP)

题目:

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 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

 

题意:

求出最长字符串长度, 左边递增右边递减两边个数相等.

思路:

从左到右求出最长递增序列,从右到左同样求出, 枚举每一个位置求出最大值,注意左右序列要相等.

AC.

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;const int maxn = 100005;int num[maxn], dp[maxn];int l[maxn], r[maxn];int main(){//freopen("in", "r", stdin);    int n;    while(~scanf("%d", &n)) {        fill(dp, dp + n, maxn);        memset(l, 0, sizeof(l));        memset(r, 0, sizeof(r));        for(int i = 0; i < n; ++i) {            scanf("%d", &num[i]);        }        for(int i = 0; i < n; ++i) {            *lower_bound(dp, dp + n, num[i]) = num[i];            l[i] = lower_bound(dp, dp+n, maxn) - dp;        }        fill(dp, dp + n, maxn);        for(int i = n - 1; i >= 0; --i) {            *lower_bound(dp, dp + n, num[i]) = num[i];            r[i] = lower_bound(dp, dp+n, maxn) - dp;        }        int ans = 1;        for(int i = 0; i < n; ++i) {            if(l[i]>1 && r[i]>1 && l[i]==r[i]) {                ans = max(ans, l[i]+r[i]-1);            }        }        printf("%d\n", ans);    }    return 0;}