Starter 188-Inversion Queries

思路讲解

two number <=L || >=R delete

FOR(i,1,Q){
	// two number <=L || >=R  delete
	cin>>L[i]>>R[i];
	if(L[i]==R[i]){
		cout<<0<<"\n";
		continue;
	}
	ll lans=ans;
	lans-=ansp.query(R[i],N);
	lans-=anss.query(1,L[i]);
	if(lans<0) cout<<"0\n";
	else cout<<lans<<"\n";
}

AC代码

https://www.codechef.com/viewsolution/1168087702

源代码

// Problem: Inversion Queries
// Contest: CodeChef - START188
// URL: https://www.codechef.com/problems/INVQUER
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)

using namespace std;

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e17)+9; // 1e18+9也是素数
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,X,K,lT,A[MAXN],L[MAXN],R[MAXN];
/*

*/
inline void init(){
	
}
struct BIT {  // 普通的单点修改区间查询树状数组，含有查找第K个空位的功能
  vector<ll> tr;
  int n;
  inline int lowbit(int x) { return x & (-x); }
  BIT(int ln) {  // 构造+初始化函数
    n = ln;
    tr.resize(n+5,0);
  }
  inline void add(int p, int x) {
  	if(p==0) return;
    while (p <= n) {
      tr[p] += x;
      p += lowbit(p);
    }
  }
  inline ll query(int l, int r) {
    ll lres = 0, rres = 0;
    l -= 1;
    while (l > 0) {
      lres += tr[l];
      l -= lowbit(l);
    }
    while (r > 0) {
      rres += tr[r];
      r -= lowbit(r);
    }
    return rres - lres;
  }
  inline int findKth(int k) {  // 找到第k个空位，当然应当是要求只能出现0和1
    int cx = 0;
    for (int i = 1 << 20; i > 0; i >>= 1) {  // 这个你可以理解为不断缩小步长的搜索
      if (cx + i <= n && lowbit(cx + i) - tr[cx + i] < k) {
        // lowbit(cx)-tr[cx+i] 表示cx+i这个树状数组区间有多少空位
        k -= lowbit(cx + i) - tr[cx + i];
        cx += i;
      }
    }
    return cx + 1;  // 我们始终让cx < k，所以返回的就是最大的比k小的，+1就是新空位出现的地方
  }
};
inline void solve(){
	cin>>N>>Q;
	init();
	FOR(i,1,N){
		cin>>A[i];
	}
	BIT trp(N),ansp(N),trs(N),anss(N);
	// 找编号比自己小但是更大的
	ll ans=0;
	FOR(i,1,N){
		ll res=trp.query(A[i],N);
		ansp.add(A[i],res);
		ans+=res;
		trp.add(A[i],1);
	}
	// index big but value small
	ROF(i,N,1){
		ll res=trs.query(1,A[i]);
		anss.add(A[i],res);
		trs.add(A[i],1);
	}
	FOR(i,1,Q){
		// two number <=L || >=R  delete
		cin>>L[i]>>R[i];
		if(L[i]==R[i]){
			cout<<0<<"\n";
			continue;
		}
		ll lans=ans;
		lans-=ansp.query(R[i],N);
		lans-=anss.query(1,L[i]);
		if(lans<0) cout<<"0\n";
		else cout<<lans<<"\n";
#ifdef LOCAL
    debug(lans);
#endif
	}
	
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>lT;
	while(lT--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://www.codechef.com/viewsolution/1168087702
*/

心路历程（WA，TLE，MLE……）

那么就是要注意这个特判

if(L[i]==R[i]){
	cout<<0<<"\n";
	continue;
}