COJ969 WZJ的数据结构（负三十一）

妈妈我会写线段树分治了！

按论文的方法，操作对应一个时间区间，将这个区间打上标记，那么操作的标记数是O(nlogn)的，询问对应一个点，将点所在的区间打上标记，那么询问的标记数也是O(nlogn)的。最后对每一条线段离线做就可以了。

#include<cstdio>
#include<cctype>
#include<queue>
#include<cstring>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define ren for(int i=first[x];i!=-1;i=next[i])
using namespace std;
inline int read() {
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
    return x*f;
}
typedef long long ll;
const int maxn=200010;
ll x0,y0,x1,y1,tmp[maxn];
int n,N,first1[maxn*3],next1[maxn*15],id1[maxn*15],ToT1;
int first2[maxn*3],next2[maxn*15],id2[maxn*15],ToT2;
struct Query {
    int tp,id;
    ll x,y;
}Q[maxn];
struct Point {
    ll x,y;int id;
    bool operator < (const Point& a) const {
        if(x!=a.x) return x>a.x;
        if(y!=a.y) return y>a.y;
    }
}A[maxn];
void pre() {
    rep(i,1,n) tmp[i]=-Q[i].y;
    sort(tmp+1,tmp+n+1);
    rep(i,1,n) Q[i].y=lower_bound(tmp+1,tmp+n+1,-Q[i].y)-tmp;
}
int ans[maxn],sumv[maxn];
void add(int x,int v) {for(;x<=n;x+=x&-x) sumv[x]+=v;}
int sum(int x) {int res=0;for(;x;x-=x&-x) res+=sumv[x];return res;}
void query1(int o,int l,int r,int ql,int qr,int v) {
    if(ql<=l&&r<=qr) {
        id1[++ToT1]=v;next1[ToT1]=first1[o];first1[o]=ToT1;
    }
    else {
        int mid=l+r>>1,lc=o<<1,rc=lc|1;
        if(ql<=mid) query1(lc,l,mid,ql,qr,v);
        if(qr>mid) query1(rc,mid+1,r,ql,qr,v);
    }
}
void query2(int o,int l,int r,int x,int v) {
    id2[++ToT2]=v;next2[ToT2]=first2[o];first2[o]=ToT2;
    if(l==r) return;
    int mid=l+r>>1,lc=o<<1,rc=lc|1;
    if(x<=mid) query2(lc,l,mid,x,v);
    else query2(rc,mid+1,r,x,v);
}
int main() {
    x0=read();y0=read();x1=read();y1=read();
    n=read();
    rep(i,1,n) {
        Q[Q[i].id=i].tp=read();
        ll x=read(),y=read();
        if(Q[i].tp==1) {
            Q[i].x=(x-x0)*(x-x0)+(y-y0)*(y-y0),Q[i].y=(x-x1)*(x-x1)+(y-y1)*(y-y1);
            query1(1,1,n,i,n,i);
        }
        else {
            Q[i].x=x*x,Q[i].y=y*y;
            query2(1,1,n,i,i);
        }
    }
    pre();
    rep(i,1,3*n) {
        int m=0;
        for(int j=first2[i];j;j=next2[j]) A[++m]=(Point){Q[id2[j]].x,Q[id2[j]].y,Q[id2[j]].id};
        for(int j=first1[i];j;j=next1[j]) A[++m]=(Point){Q[id1[j]].x,Q[id1[j]].y,0};
        sort(A+1,A+m+1);
        rep(j,1,m) if(!A[j].id) add(A[j].y,1); else ans[A[j].id]+=sum(A[j].y-1);
        rep(j,1,m) if(!A[j].id) add(A[j].y,-1);
    }
    rep(i,1,n) if(Q[i].tp==2) printf("%d\n",ans[i]);
    return 0;
}

CDQ分治常数比较小：

#include<cstdio>
#include<cctype>
#include<queue>
#include<cstring>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define ren for(int i=first[x];i!=-1;i=next[i])
using namespace std;
inline int read() {
    int x=0,f=1;char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c==‘-‘) f=-1;
    for(;isdigit(c);c=getchar()) x=x*10+c-‘0‘;
    return x*f;
}
typedef long long ll;
const int maxn=200010;
ll x0,y0,x1,y1,tmp[maxn];
struct Query {
    int tp,id;
    ll x,y;
}Q[maxn];
struct Point {
    ll x,y;int id;
    bool operator < (const Point& a) const {
        if(x!=a.x) return x>a.x;
        if(y!=a.y) return y>a.y;
    }
}A[maxn];
int n,ans[maxn],sumv[maxn];
void add(int x,int v) {for(;x<=n;x+=x&-x) sumv[x]+=v;}
int sum(int x) {int res=0;for(;x;x-=x&-x) res+=sumv[x];return res;}
void solve(int L,int R) {
    if(L>=R) return;
    int mid=L+R>>1,m0=0,m=0;
    solve(L,mid);solve(mid+1,R);
    rep(i,L,mid) if(Q[i].tp==1) A[++m0]=(Point){Q[i].x,Q[i].y,0};
    if(!m0) return;m=m0;
    rep(i,mid+1,R) if(Q[i].tp==2) A[++m0]=(Point){Q[i].x,Q[i].y,Q[i].id};
    if(m0==m) return;m=m0;
    sort(A+1,A+m+1);
    rep(i,1,m) if(!A[i].id) add(A[i].y,1); else ans[A[i].id]+=sum(A[i].y-1);
    rep(i,1,m) if(!A[i].id) add(A[i].y,-1);
}
void pre() {
    rep(i,1,n) tmp[i]=-Q[i].y;
    sort(tmp+1,tmp+n+1);
    rep(i,1,n) Q[i].y=lower_bound(tmp+1,tmp+n+1,-Q[i].y)-tmp;
}
int main() {
    x0=read();y0=read();x1=read();y1=read();
    n=read();
    rep(i,1,n) {
        Q[Q[i].id=i].tp=read();
        ll x=read(),y=read();
        if(Q[i].tp==1) Q[i].x=(x-x0)*(x-x0)+(y-y0)*(y-y0),Q[i].y=(x-x1)*(x-x1)+(y-y1)*(y-y1);
        else Q[i].x=x*x,Q[i].y=y*y;
    }
    pre();solve(1,n);
    rep(i,1,n) if(Q[i].tp==2) printf("%d\n",ans[i]);
    return 0;
}