BZOJ 3289 Mato的文件管理

题解：莫队，离散化后用树状数组维护

/**************************************************************
    Problem: 3289
    User: ws_zzyer
    Language: C++
    Result: Accepted
    Time:4904 ms
    Memory:2856 kb
****************************************************************/
 
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=50009;
const int SIZ=300;
int n,m;
int a[maxn];
 
int b[maxn],nn; 
int c[maxn]; 
inline int lowbit(int x){
    return x&(-x);
}
void Addp(int x,int val){
    while(x<=nn){
        c[x]+=val;
        x+=lowbit(x);
    }
}
int Querysum(int x){
    int ret=0;
    while(x){
        ret+=c[x];
        x-=lowbit(x);
    }
    return ret;
}
 
int p[maxn];
int qx[maxn],qy[maxn];
long long ans[maxn];
bool cmp(const int &rhs1,const int &rhs2){
    if((qx[rhs1]-1)/SIZ==(qx[rhs2]-1)/SIZ){
        return qy[rhs1]<qy[rhs2];
    }else{
        return qx[rhs1]<qx[rhs2];
    }
}
 
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;++i){
        int x;scanf("%d",&x);
        a[i]=b[i]=x;
    }
    sort(b+1,b+1+n);
    nn=unique(b+1,b+1+n)-b-1;
    for(int i=1;i<=n;++i)a[i]=lower_bound(b+1,b+1+nn,a[i])-b;
     
    scanf("%d",&m);
    for(int i=1;i<=m;++i){
        scanf("%d%d",&qx[i],&qy[i]);
    }
    for(int i=1;i<=m;++i)p[i]=i;
    sort(p+1,p+1+m,cmp);
     
    int l=1,r=0;
    long long nowans=0;
    for(int i=1;i<=m;++i){
        int t=p[i];
        int x=qx[t];
        int y=qy[t];
        while(l<x){
            Addp(a[l],-1);
            nowans-=Querysum(a[l]-1);
            ++l;
        }
        while(l>x){
            --l;
            Addp(a[l],1);
            nowans+=Querysum(a[l]-1);
        }
        while(r<y){
            ++r;
            Addp(a[r],1);
            nowans+=(r-l+1-Querysum(a[r]));
        }
        while(r>y){
            Addp(a[r],-1);
            nowans-=(r-l-Querysum(a[r]));
            --r;
        }
//      cout<<l<<' '<<r<<endl;
        ans[t]=nowans;
    }
    for(int i=1;i<=m;++i)printf("%lld
",ans[i]);
    return 0;
}