每次询问区间[l,r]内相邻交换,排好序的最小交换次数
这个东西等于区间内的逆序对数(易证)
然后用莫队就行了,树状数组每次O(logn)求出区间内比这个数大或小的数
总复杂度O(nn√logn)
code:
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<cmath>
#include<ctime>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<complex>
#include<iostream>
#include<algorithm>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;
inline void read(int &x)
{
char c;
while(!((c=getchar())>='0'&&c<='9'));
x=c-'0';
while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';
}
const int maxn = 51000;
struct A
{
int c,i;
}c[maxn];
inline bool cmpc(const A x,const A y) {return x.c<y.c;}
int a[maxn];
int s[maxn],n;
void upd(int x,const int c){for(;x<=n;x+=lowbit(x))s[x]+=c;}
int query(int x){int re=0;for(;x;x-=lowbit(x))re+=s[x];return re;}
struct node
{
int l,r,i;
node(){}
node(const int _l,const int _r,const int _i){l=_l;r=_r;i=_i;}
}q[maxn]; int m;
int N,id[maxn],L[maxn];
inline bool cmp(node x,node y){return id[x.l]==id[y.l]?x.r<y.r:x.l<y.l;}
ll ans[maxn];
void solve()
{
sort(q+1,q+m+1,cmp);
int pos=1; int l=1,r=0;
ll tmp=0ll,s=0ll;
for(int i=1;i<=id[n];i++)
{
while(pos<=m&&q[pos].l<L[i+1])
{
while(r<q[pos].r) tmp+=(s++)-(ll)query(a[++r]),upd(a[r],1);
while(l>q[pos].l) tmp+=(ll)query(a[--l]-1),upd(a[l],1),s++;
while(r>q[pos].r) tmp-=s-(ll)query(a[r]),upd(a[r--],-1),s--;
while(l<q[pos].l) tmp-=(ll)query(a[l]-1),upd(a[l++],-1),s--;
ans[q[pos++].i]=tmp;
}
}
}
int main()
{
read(n); N=sqrt(n);
for(int i=1;i<=n;i++) id[i]=(i-1)/N+1; id[n+1]=id[n]+1;
for(int i=1;i<=id[n];i++) L[i]=(i-1)*N+1; L[id[n]+1]=n+1;
for(int i=1;i<=n;i++) read(c[i].c),c[i].i=i,a[i]=c[i].c;
sort(c+1,c+n+1,cmpc); c[0].c=c[1].c-1;
for(int i=1,k=0;i<=n;i++)
{
if(c[i].c!=c[i-1].c) k++;
a[c[i].i]=k;
}
read(m);
for(int i=1;i<=m;i++)
{
int l,r; read(l); read(r);
q[i]=node(l,r,i);
}
solve();
for(int i=1;i<=m;i++) printf("%lld\n",ans[i]);
return 0;
}