Codeforces round #649 B. Most socially-distanced subsequence

题目链接

题意：让求|s1-s2|+|s2-s3|+…|sk-1 - sk|最大时的最短序列。

思路：去掉绝对值之后我们发现如果每三个数是有序的那么我们可以消去中间的数不影响答案，看能消多少。

#include 
#include 
#include 
#include 
#include
#include
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
typedef pair<int,int> PII;
const int mod=1e4+7;
const int N=2e5+10;
const int inf=0x7f7f7f7f;


ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b)
{
    return a*(b/gcd(a,b));
}

template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-')
            op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op)
        x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0)
        x = -x, putchar('-');
    if(x >= 10)
        write(x / 10);
    putchar('0' + x % 10);
}
int vis[N];
int a[N];

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(vis,0,sizeof vis);
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        int ans=n;
        for(int i=2;i<=n-1;i++)
        {
            if((a[i]<=a[i-1]&&a[i]>=a[i+1])||(a[i]>=a[i-1]&&a[i]<=a[i+1]))
            {
                vis[i]++;ans--;
            }
        }
        cout<<ans<<endl;
        for(int i=1;i<=n;i++)
        {
            if(!vis[i])
                cout<<a[i]<<endl;
        }
        cout<<endl;
    }

    return 0;
}