Codeforces Round #649 (Div. 2) B. Most socially-distanced subsequence (数学,差分)

• 题意:有一长度为\(n\)的数组,求一子序列,要求子序列中两两差的绝对值最大,并且子序列尽可能短.

• 题解:将数组看成坐标轴上的点,其实就是求每个单调区间的端点,用差分数组来判断单调性.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  
  int t;
  int n,a[N];
  int b[N];
  vector<int> v;
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
  	cin>>t;
  	 while(t--){
  	 	cin>>n;
  	 	v.clear();
  	 	for(int i=1;i<=n;++i){
  	 		cin>>a[i];
  	 		b[i]=a[i]-a[i-1];
  	 	}
  	 	v.pb(a[1]);
  	 	for(int i=3;i<=n;++i){
  	 		if((b[i]>0&&b[i-1]<0)||(b[i]<0&&b[i-1]>0)){
  	 			v.pb(a[i-1]);
  	 		}
  	 	}
  	 	v.pb(a[n]);
  	 	cout<<v.size()<<endl;
  	 	for(auto w:v){
  	 		cout<<w<<" ";
  	 	}
  	 	cout<<"\n";
  	 }
  
      return 0;
  }