BNUOJ 14381 Wavio Sequence

Wavio Sequence

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 10534
64-bit integer IO format: %lld      Java class name: Main

 

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

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

Sample Output

9

9

1

解题：最长上升子序列加强版。单调队列优化！！！重点。先顺着求最长上升子序列，再逆着求最长上升子序列。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 10100;
int dp1[maxn],dp2[maxn],d[maxn],q[maxn];
int bsearch(int lt,int rt,int val) {
    while(lt <= rt) {
        int mid  = (lt+rt)>>1;
        if(q[mid] < val) lt = mid+1;//严格上升单调取少于符号，上升的取少于等于
        else rt = mid-1;
    }
    return lt;
}
int main() {
    int n,i,j,head,tail;
    while(~scanf("%d",&n)) {
        for(i = 0; i < n; i++)
            scanf("%d",d+i);
        head = tail = 0;
        for(i = 0; i < n; i++) {
            if(head == tail) {
                q[head++] = d[i];
                dp1[i] = head-tail;
            }else if(d[i] > q[head-1]){
                q[head++] = d[i];
                dp1[i] = head-tail;
            }else{
                int it = bsearch(tail,head-1,d[i]);
                dp1[i] = it - tail + 1;
                q[it] = d[i];
            }
        }
        head = tail = 0;
        for(i = n-1; i >= 0; i--) {
            if(head == tail) {
                q[head++] = d[i];
                dp2[i] = head-tail;
            }else if(d[i] > q[head-1]){
                q[head++] = d[i];
                dp2[i] = head-tail;
            }else{
                int it = bsearch(tail,head-1,d[i]);
                dp2[i] = it - tail + 1;
                q[it] = d[i];
            }
        }
        int ans = 1;
        for(i = 0; i < n; i++){
            ans = max(ans,min(dp1[i],dp2[i])*2-1);
        }
        printf("%d\n",ans);
    }
    return 0;
}