B. Most socially-distanced subsequence (thinking + simulation)

B. Most socially-distanced subsequence

Question:
Give an array $a$ , find a subsequencessin the array$s$ (length is$x$)，满足“${\sum }_{i=1}^x|s[i]-s[i-1]|$ . Maximum "If there is more than one answer lies$x$ smaller subsequence

Idea:
We observe an increasing sequence s = {1, 2, 3, 4, 6} s = \{1, 2, 3, 4, 6\}s={ 1，2，3，4，6}，则：
${\sum }_{i=1}^x|s[i]-s[i-1]|=5$，因为$s[i]>s[i-1]$ , so when we disassemble the formula, we find that only$s[5]-s[1]=6-1=5.$ Similarly, when the sequence is decreasing, the same conclusion can be obtained. (Solved the shortest problem).

Sequence{x, z, y} \{x,z,y\}{ x,with ,y } , if on the number line$z$在$between\; xandy$ (such as${with}_1$) ，$|y-z|+|z-x|=and-x$ if it is like${with}_2$In the same way, we can find $|y-z|+|z-x|=with-and+with-x$ , this finds${\sum }_{i=1}^x|s[i]-s[i-1]|$ largest andxxThe methods used to minimize$x$ are the same.

The rest is our simulation, to find out the continuous increasing or decreasing sequence in the array a, and remove the middle element without affecting the result.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+10;
const int inf = 0x7ffffff;
const ll INF  = 0x7fffffffffff;
ll f[][2] = {
    
    1, 0, 0, 1, -1, 0, 0, -1}, n, m;
ll s[N];
int read() {
    
    
    int x = 0; int f = 1; char s = getchar();
    while(s < '0' || s > '9') {
    
    if(s == '-') f = -1; s = getchar();}
    while(s >= '0' && s <= '9') {
    
    x = (x << 3) + (x << 1) + s - 48; s = getchar();}
    return x * f;
}
int main() {
    
    
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    int t;
    cin >> t;
    while(t--) {
    
    
        n = read();
        m = n;
        for(int i=1; i<=n; i++) s[i] = read();
        if(n == 2) {
    
    
            cout << 2 << endl;
            for(int i=1; i<=n; i++) cout << s[i] << ' ';
            cout << endl; 
            continue;
        }
        int i = 1, j = 2, k = 3;//从数组下标1，2，3开始扫描，每三个三个比较。
        while(k <= n) {
    
    
        	//判断是否连续递增或
            if(s[j]>s[i] && s[k]>s[j] || s[j]<s[i] && s[j]>s[k]) {
    
    
                m--;
                s[j] = -1;/做标记。
                j++, k++;
            }
            else i = j, j++, k++;
        }
        cout << m << endl;
        for(int i=1; i<=n; i++) {
    
    
            if(s[i] != -1) cout << s[i] << ' ';
        }
        cout << endl;
    }
    return 0;
}