Most socially-distanced subsequence（贪心）

题目

Given a permutation p of length n, find its subsequence s1, s2, …, sk of length at least 2 such that:

|s1−s2|+|s2−s3|+…+|sk−1−sk| is as big as possible over all subsequences of p with length at least 2.
Among all such subsequences, choose the one whose length, k, is as small as possible.
If multiple subsequences satisfy these conditions, you are allowed to find any of them.

A sequence a is a subsequence of an array b if a can be obtained from b by deleting some (possibly, zero or all) elements.

A permutation of length n is an array of length n in which every element from 1 to n occurs exactly once.

Input
The first line contains an integer t ( 1 ≤ t ≤ 2 × 1 0 4 ) t (1≤t≤2\times10^4) t(1≤t≤2×104)— the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n ( 2 ≤ n ≤ 1 0 5 ) n (2≤n≤10^5) n(2≤n≤105) — the length of the permutation p.

The second line of each test case contains n integers p1, p2, …, pn (1≤pi≤n, pi are distinct) — the elements of the permutation p.

The sum of n across the test cases doesn’t exceed 105.

Output
For each test case, the first line should contain the length of the found subsequence, k. The second line should contain s1, s2, …, sk — its elements.

If multiple subsequences satisfy these conditions, you are allowed to find any of them.

Example
Input
2
3
3 2 1
4
1 3 4 2
Output
2
3 1
3
1 4 2

思路

把这个排列想象成一个个长条排放在坐标轴上，如1，3，2 ，4
如此看来，不难看出要让子数组相邻数字的差的绝对值的和最大要递增或递减地去寻找，从第一个开始，放入子数组的第一位，往后若递增（a[i]>a[i-1]），则一直往下寻找，直到出现转折（a[i]<a[i-1]），再把a[i-1]放到子数组的下一个。若往后递减也类似如此去寻找子数组里的各个数字。在这里插入代码片

代码

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
using namespace std;const int N=1e5+5;int a[N];
int sub[N];int main()
{int t;scanf("%d",&t);int n;while(t--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&a[i]);}int temp=0;sub[temp++]=a[0];int i=1;while(i<n){if(a[i]<a[i-1]){while(a[i]<a[i-1] && i<n){i++;}sub[temp++]=a[i-1];}else{while(a[i]>a[i-1] && i<n){i++;}sub[temp++]=a[i-1];}}printf("%d\n",temp);for(int j=0;j<temp;j++){printf("%d ",sub[j]);}printf("\n");}return 0;
}

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Most socially-distanced subsequence（贪心）

题目

思路

代码

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