BZOJ3289: Mato的文件管理

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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;}

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