BZOJ4170: 极光

将每个位置的位置定为x，graze值定为y
那么任意位置上出现过的任意数值就对应上了平面上的若干个点
每次询问和一个点的哈密顿距离 <= x的点的个数
将哈密顿距离转为切比雪夫距离(x,y)−>(x+y,x−y)
和一个点切比雪夫距离 <= x的点对应平面上一个矩形
那么询问就变成了询问平面上一个矩形内的点的个数
用KD-tree或cdq都是兹瓷的

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 longusing namespace std;inline void up(int &x,const int &y){if(x<y)x=y;}inline void down(int &x,const int &y){if(x>y)x=y;}inline void read(int &x){    char c; int f=1;    while(!((c=getchar())>='0'&&c<='9')) if(c=='-') f*=-1;    x=c-'0';    while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';    x*=f;}const int maxn = 161000;const int maxm = 101000;const int maxd = 2;struct KD_tree{    int i,fa,lc,rc,pos[maxd],u[maxd],d[maxd],c,s;}kd[maxn]; int N,root;int id[maxn];int cmp_d; inline bool cmp(KD_tree x,KD_tree y){return x.pos[cmp_d]<y.pos[cmp_d];}inline void push_up(const int x){    int lc=kd[x].lc,rc=kd[x].rc;    if(lc)     {        for(int i=0;i<maxd;i++)             up(kd[x].u[i],kd[lc].u[i]),            down(kd[x].d[i],kd[lc].d[i]);        kd[x].c+=kd[lc].c;    }    if(rc)    {        for(int i=0;i<maxd;i++)            up(kd[x].u[i],kd[rc].u[i]),            down(kd[x].d[i],kd[rc].d[i]);        kd[x].c+=kd[rc].c;    }}int build_(const int l,const int r,const int ff,int D){    int x=(l+r)>>1;    cmp_d=D; std::nth_element(kd+l,kd+x,kd+r+1,cmp);    for(int i=0;i<maxd;i++) kd[x].u[i]=kd[x].d[i]=kd[x].pos[i];    kd[x].c=kd[x].s; kd[x].fa=ff;    if(l!=x) kd[x].lc=build_(l,x-1,x,!D);    if(x!=r) kd[x].rc=build_(x+1,r,x,!D);    push_up(x);    return x;}void upd(int x){    kd[x].c++;    while(x=kd[x].fa) kd[x].c++;}int u[maxd],d[maxd];int query(const int x){    for(int i=0;i<maxd;i++)        if(kd[x].u[i]<d[i]||kd[x].d[i]>u[i]) return 0;    if(d[0]<=kd[x].d[0]&&kd[x].u[0]<=u[0]&&d[1]<=kd[x].d[1]&&kd[x].u[1]<=u[1]) return kd[x].c;    int re=0;    if(d[0]<=kd[x].pos[0]&&kd[x].pos[0]<=u[0]&&d[1]<=kd[x].pos[1]&&kd[x].pos[1]<=u[1]) re+=kd[x].s;    return re+query(kd[x].lc)+query(kd[x].rc);}struct point{int x,y;};inline bool operator <(point x,point y){return x.x==y.x?x.y<y.y:x.x<y.x;}map<point,int>h;struct node{    int x,y,k,q;    node(){}    node(const int _x,const int _y,const int _k,const int _q){x=_x;y=_y;k=_k;q=_q;}}q[maxm];int n,m;int v[maxn];int main(){    read(n); read(m);    for(int i=1;i<=n;i++)    {        int x; read(x); v[i]=x;        point tmp; tmp.x=i+x,tmp.y=i-x;        if(h.count(tmp)>0) kd[h[tmp]].s++;        else        {            kd[h[tmp]=++N].s=1;            kd[N].pos[0]=tmp.x,kd[N].pos[1]=tmp.y;            kd[N].i=N;        }    }    char str[10];    for(int i=1;i<=m;i++)    {        scanf("%s",str);        int x,y; read(x); read(y);        if(str[0]=='M')        {            point tmp; tmp.x=x+y,tmp.y=x-y;            q[i]=node(tmp.x,tmp.y,0,0);            if(h.count(tmp)) q[i].k=h[tmp];            else            {                kd[h[tmp]=q[i].k=++N].s=0;                kd[N].pos[0]=tmp.x,kd[N].pos[1]=tmp.y;                kd[N].i=N;            }            v[x]=y;        }        else q[i]=node(x+v[x],x-v[x],y,1);    }    root=build_(1,N,0,0);    for(int i=1;i<=N;i++) id[kd[i].i]=i;    for(int i=1;i<=m;i++)    {        if(q[i].q==0)        {            q[i].k=id[q[i].k];            kd[q[i].k].s++;            upd(q[i].k);        }        else        {            d[0]=q[i].x-q[i].k,u[0]=q[i].x+q[i].k;            d[1]=q[i].y-q[i].k,u[1]=q[i].y+q[i].k;            point tmp; tmp.x=q[i].x,tmp.y=q[i].y;            printf("%d\n",query(root));        }    }    return 0;}