BZOJ 2683/4066(简单题-kd-tree重构)

初始(N*N)矩阵全部为0，维护2个操作：
1.矩阵加
2.矩阵求和
1<=N<=500000,操作数不超过200000个，内存限制20M

2683离线
4066强制在线，数据加强

对于离线，我们可以离散化，然后树套树，cdq，怎么开心怎么来
但是在线的情况，我们不能离散化，于是用kd-tree树上求和
Bzoj 2683

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (1000000000)#define F (100000007)#define MAXN (200000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n;int cmp_d=0;class node{public:    int x[2];    int l,r,minv[2],maxv[2];    int w,sumv;    node(){}    node(int a,int b,int _w){MEM(x) l=r=0; w=sumv=_w; x[0]=a,x[1]=b; Rep(i,2) minv[i]=maxv[i]=x[i];}    int& operator[](int i){return x[i]; } };int cmp(node a,node b){return a[cmp_d]<b[cmp_d];    }class KD_Tree{public:     node a[MAXN*3];    KD_Tree()    {       }    void mem()    {    }    void update(node& o)    {        o.sumv=o.w;        if (o.l)        {            node p=a[o.l];            Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);            Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);                o.sumv+=p.sumv;        }        if (o.r)        {            node p=a[o.r];            Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);            Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);                o.sumv+=p.sumv;        }    }    int build(int L,int R,int nowd)    {        int m=(L+R)>>1;        cmp_d=nowd;        nth_element(a+L+1,a+m+1,a+R+1,cmp);        if (L^m) a[m].l=build(L,m-1,nowd^1);        if (R^m) a[m].r=build(m+1,R,nowd^1);        update(a[m]);        return m;    }     int root;    void _build(int L,int R,int nowd) //1-n的节点 至少为1     {        root=build(L,R,nowd);    }    void insert(int o,int k,int nowd)    {        int p=a[o].x[nowd];        int p2=a[k].x[nowd];        if (p2<=p)         {            if (a[o].l)                 insert(a[o].l,k,nowd^1);            else a[o].l=k;        }        else        {            if (a[o].r)                insert(a[o].r,k,nowd^1);                    else a[o].r=k;        }           update(a[o]);       }    void _insert(int k,int nowd)    {        int p=root;        insert(root,k,nowd);            }    int _x1,_y1,_x2,_y2;    int _ans;    void ask(int o)    {        if (o==0) return;        if (_x1<=a[o].minv[0] && a[o].maxv[0]<=_x2 && _y1<=a[o].minv[1] && a[o].maxv[1]<=_y2 ) {            _ans+=a[o].sumv;return;        }               if (_x1<=a[o].x[0] && a[o].x[0]<=_x2 && _y1<=a[o].x[1] && a[o].x[1]<=_y2 ) {            _ans+=a[o].w;        }               if (a[o].l) {            int p=a[o].l;            if (a[p].minv[0]<=_x2 && _x1<=a[p].maxv[0] && a[p].minv[1]<=_y2 && _y1<=a[p].maxv[1] )                 ask(p);        }        if (a[o].r) {            int p=a[o].r;            if (a[p].minv[0]<=_x2 && _x1<=a[p].maxv[0] && a[p].minv[1]<=_y2 && _y1<=a[p].maxv[1] )                 ask(p);        }    }    int _ask(int x1,int y1,int x2,int y2)    {        _x1=x1;_y1=y1;_x2=x2;_y2=y2;        ;_ans=0;        ask(root);        return _ans;     }}S;int main(){    int N;    cin>>N;    n=0;    S.a[++n]=node(0,0,0);    S._build(1,n,0);    int p;    int ans=0;    while (scanf("%d",&p)==1 && p^3)    {        if (p==1) {            int x,y,A;            scanf("%d%d%d",&x,&y,&A);            S.a[++n]=node(x,y,A);            S._insert(n,0);        } else {            int x1,y1,x2,y2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            ans=S._ask(x1,y1,x2,y2);            printf("%d\n",ans);         }    }     return 0;}

但是这样不能保证kd-tree的平衡，过不了4066，
于是我们对kd-tree重构

Bzoj 4066

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (1000000000)#define F (100000007)#define MAXN (200000+10)#define fac 0.65typedef long long ll;int n;int cmp_d=0;class node{public:    int x[2];    int l,r,minv[2],maxv[2];    int w,sumv,siz;    node(){}    node(int a,int b,int _w){l=r=0; siz=1; w=sumv=_w; x[0]=a,x[1]=b; Rep(i,2) minv[i]=maxv[i]=x[i];}    int& operator[](int i){return x[i]; } };int cmp(node a,node b){return a[cmp_d]<b[cmp_d];    }int cmp2(int i,int j); int p;char c;int read(){    while (c=getchar(),!isdigit(c));    p=c-'0';    while (isdigit(c=getchar())) p=p*10+c-'0'; return p;}class KD_Tree{public:     node a[MAXN];    void update(node& o)    {        o.sumv=o.w;        o.minv[0]=o.maxv[0]=o.x[0];o.minv[1]=o.maxv[1]=o.x[1];         o.siz=1;        if (o.l)        {            node p=a[o.l];            Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);            Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);                o.sumv+=p.sumv;            o.siz+=p.siz;        }        if (o.r)        {            node p=a[o.r];            Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);            Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);                o.sumv+=p.sumv;            o.siz+=p.siz;        }    }    int build(int L,int R,int nowd,node *a)    {        int m=(L+R)>>1;        cmp_d=nowd;        nth_element(a+L+1,a+m+1,a+R+1,cmp);        if (L^m) a[m].l=build(L,m-1,nowd^1,a);        if (R^m) a[m].r=build(m+1,R,nowd^1,a);        update(a[m]);        return m;    }     int po[MAXN],pt;    void dfs(int x)    {        po[++pt]=x;        if (a[x].l) dfs(a[x].l);        if (a[x].r) dfs(a[x].r);    }    int rebuild(int L,int R,int nowd)    {        int m=(L+R)>>1;        cmp_d=nowd;        nth_element(po+L+1,po+m+1,po+R+1,cmp2);        int now=po[m];        a[now].l=a[now].r=0;        if (L^m) a[now].l=rebuild(L,m-1,nowd^1);        if (R^m) a[now].r=rebuild(m+1,R,nowd^1);        update(a[now]);        return now;    }     int root;    void _build(int L,int R,int nowd) //1-n的节点 至少为1     {        root=build(L,R,nowd,a);    }    int insert(int o,int k,int nowd)    {        if (!o) return k;        int p=a[o].x[nowd];        int p2=a[k].x[nowd];        int nx=0;        if (p2<=p)         {            a[o].l=insert(a[o].l,k,nowd^1);            nx=a[o].l;        }        else        {            a[o].r=insert(a[o].r,k,nowd^1);            nx=a[o].r;        }           update(a[o]);           if (a[nx].siz>(double)a[o].siz*fac)        {            pt=0,dfs(o);            o=rebuild(1,pt,nowd);           }        return o;    }    void _insert(int k,int nowd)    {        int p=root;        root = insert(root,k,nowd);         }    int _x1,_y1,_x2,_y2;    int _ans;    void ask(int o)    {        if (o==0) return;        if (_x1<=a[o].minv[0] && a[o].maxv[0]<=_x2 && _y1<=a[o].minv[1] && a[o].maxv[1]<=_y2 ) {            _ans+=a[o].sumv;return;        }               if (_x1<=a[o].x[0] && a[o].x[0]<=_x2 && _y1<=a[o].x[1] && a[o].x[1]<=_y2 ) {            _ans+=a[o].w;        }               if (a[o].l) {            int p=a[o].l;            if (a[p].minv[0]<=_x2 && _x1<=a[p].maxv[0] && a[p].minv[1]<=_y2 && _y1<=a[p].maxv[1] )                 ask(p);        }        if (a[o].r) {            int p=a[o].r;            if (a[p].minv[0]<=_x2 && _x1<=a[p].maxv[0] && a[p].minv[1]<=_y2 && _y1<=a[p].maxv[1] )                 ask(p);        }    }    int _ask(int x1,int y1,int x2,int y2)    {        _x1=x1;_y1=y1;_x2=x2;_y2=y2;        ;_ans=0;        ask(root);        return _ans;     }}S;int cmp2(int i,int j){return S.a[i].x[cmp_d]<S.a[j].x[cmp_d];   }int main(){//  freopen("bzoj4066_data.in","r",stdin);    int N=read();    n=0;    S.a[++n]=node(N/2,N/2,0);    S._build(1,n,0);    int p;    int ans=0;    int x,y,A;    int x1,y1,x2,y2;    while (scanf("%d",&p)==1 && p^3)    {//      cout<<"t"<<endl;        if (p==1) {            x=read(),y=read(),A=read();            x^=ans;y^=ans;A^=ans;            S.a[++n]=node(x,y,A);            S._insert(n,0);        } else {            x1=read(),y1=read(),x2=read(),y2=read();            x1^=ans,y1^=ans,x2^=ans,y2^=ans;            ans=S._ask(x1,y1,x2,y2);            printf("%d\n",ans);         }    }     return 0;}