Codeforces Round #744 (Div. 3) E2. Array Optimization by Deque (贪心,逆序对)

• 题意:有一长度为\(n\)的序列,正向遍历,对于第\(i\)个元素,可以将其插入deque的队头或者队尾,问你最终得到deque后,逆序对最少是多少?

• 题解:假如将当前这个数插入队头,那么新增的逆序对就是\([2,len]\)中小于\(a[i]\)的个数,插入队尾也是同理,结合逆序对的求法,我们可以用线段树分别求出插入队头和队尾的贡献,取最小,然后update,具体看代码吧.

• 题解:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 2e5 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
   
  int n;
  vector<ll> all;
  struct Node{
      int l,r;
      int cnt;
  }tr[N<<4];
   
  int get(int x){
      return lower_bound(all.begin(),all.end(),x)-all.begin()+1;
  }
   
  void push_up(int u){
      tr[u].cnt=tr[u<<1].cnt+tr[u<<1|1].cnt;
  }
   
  void build(int u,int l,int r){
      if(l==r){
          tr[u]={l,r,0};
          return;
      }
      tr[u]={l,r,0};
      int mid=(l+r)>>1;
      build(u<<1,l,mid);
      build(u<<1|1,mid+1,r);
      push_up(u);
  }
   
  void update(int u,int x){
      if(tr[u].l==tr[u].r){
          tr[u].cnt++;
          return;
      }
      int mid=(tr[u].l+tr[u].r)>>1;
      if(x<=mid) update(u<<1,x);
      else update(u<<1|1,x);
      push_up(u);
  }
   
  ll query(int u,int L,int R){
     if(tr[u].l>=L && tr[u].r<=R){
         return tr[u].cnt;
     } 
     int mid=(tr[u].l+tr[u].r)>>1;
     ll res=0;
     if(L<=mid) res+=query(u<<1,L,R);
     if(R>mid) res+=query(u<<1|1,L,R);
     return res;
  }
   
  int main() {
      int _;
      scanf("%d",&_);
      while(_--){
          all.clear();
          scanf("%d",&n);
          vector<ll> a(n+1);
          for(int i=1;i<=n;++i){
              scanf("%lld",&a[i]);
              all.pb(a[i]);
          }
          sort(all.begin(),all.end());
          all.erase(unique(all.begin(),all.end()),all.end());
          ll ans=0;
          build(1,0,(int)all.size()+1);
          for(int i=1;i<=n;++i){
              int now=get(a[i]);
              ll sum_l=query(1,0,now-1),sum_r=query(1,now+1,(int)all.size()+1);
              ans+=min(sum_l,sum_r);
              update(1,now);
          }
          printf("%lld\n",ans);
      }
      return 0;
  }
  

标签：Deque,744,int,ll,队头,Codeforces,const,逆序,define
来源： https://www.cnblogs.com/lr599909928/p/15355086.html