题意:单点更新,大矩阵(\(n*n,n≤10^5\))求和
二维的KD树能使最坏情况不高于\(O(N\sqrt{N})\)
核心在于query时判断当前子树维护的区间是否有交集/当前子节点是否在块中,然后暴力....
然后跑30s的KD树...
维护size的重构方法不知为何T了,借鉴了hzwer的按数目重构
一个微小的工作是如果更新时点相同那就直接合并,不要另开(虽然还是可以轻易地被卡)
T了三天三夜的题总算告一段落了
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 2e5+11;
const int INF = 0x7fffffff;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int D;
struct Point{
int x[2],val;
bool operator < (const Point &rhs) const{
return x[D]<rhs.x[D];
}
};
//struct REC{
// int x[2],val;
// bool operator < (const REC &rhs) const{
// return x[D]<rhs.x[D];
// }
//}rec[MAXN];
struct KD{
int son[MAXN][2],lx[MAXN][2],rx[MAXN][2];
int size[MAXN];
ll sum[MAXN];
Point p[MAXN];
int root,tot;
void pu(int o){
int lc=son[o][0],rc=son[o][1];
rep(i,0,1){
lx[o][i]=rx[o][i]=p[o].x[i];
if(lc){
if(lx[lc][i]<lx[o][i]) lx[o][i]=lx[lc][i];
if(rx[lc][i]>rx[o][i]) rx[o][i]=rx[lc][i];
}
if(rc){
if(lx[rc][i]<lx[o][i]) lx[o][i]=lx[rc][i];
if(rx[rc][i]>rx[o][i]) rx[o][i]=rx[rc][i];
}
}
sum[o]=p[o].val+sum[lc]+sum[rc];
size[o]=1+size[lc]+size[rc];
}
void init(){
root=tot=D=0;
}
bool ok(int o){
int lc=son[o][0],rc=son[o][1];
int ls=size[lc]<<2;
int rs=size[rc]<<2;
int sz=size[o]<<1|1;
return ls<=sz&&rs<=sz;
}
void insert(int &o,int now,int x,int y,int v){
if(!o){
o=++tot;
p[o].x[0]=lx[o][0]=rx[o][0]=x;
p[o].x[1]=lx[o][1]=rx[o][1]=y;
p[o].val=sum[o]=v;size[o]=1;
son[o][0]=son[o][1]=0;
return;
}else if(p[o].x[0]==x&&p[o].x[1]==y){
p[o].val+=v;sum[o]+=v;
return;
}else if(now==0){
if(x<p[o].x[0]) insert(son[o][0],1,x,y,v);
else insert(son[o][1],1,x,y,v);
}else{
if(y<p[o].x[1]) insert(son[o][0],0,x,y,v);
else insert(son[o][1],0,x,y,v);
}
pu(o);
}
bool inside(int x_1,int y_1,int x_2,int y_2,int X1,int Y1,int X2,int Y2){
return x_1<=X1&&X2<=x_2&&y_1<=Y1&&Y2<=y_2;
}
bool outside(int x_1,int y_1,int x_2,int y_2,int X1,int Y1,int X2,int Y2){
return x_1>X2||x_2<X1||y_1>Y2||y_2<Y1;
}
ll query(int o,int x_1,int y_1,int x_2,int y_2){
if(!o) return 0;
#define RECT lx[o][0],lx[o][1],rx[o][0],rx[o][1]
#define POT p[o].x[0],p[o].x[1],p[o].x[0],p[o].x[1]
if(inside(x_1,y_1,x_2,y_2,RECT)) return sum[o];
if(outside(x_1,y_1,x_2,y_2,RECT)) return 0;
ll res=0;
if(inside(x_1,y_1,x_2,y_2,POT)) res=p[o].val;
return res+query(son[o][0],x_1,y_1,x_2,y_2)+query(son[o][1],x_1,y_1,x_2,y_2);
}
int rebuild(int now,int l,int r){
int mid=l+r>>1; son[mid][0]=son[mid][1]=0;
D=now; nth_element(p+l,p+mid,p+r+1);
// p[mid].x[0]=rec[mid].x[0];p[mid].x[1]=rec[mid].x[1];
// val[mid]=sum[mid]=rec[mid].val;size[mid]=1;
sum[mid]=p[mid].val; size[mid]=1;
lx[mid][0]=rx[mid][0]=p[mid].x[0];
lx[mid][1]=rx[mid][1]=p[mid].x[1];
if(l<mid) son[mid][0]=rebuild(now^1,l,mid-1);
if(r>mid) son[mid][1]=rebuild(now^1,mid+1,r);
pu(mid);
return mid;
}
void recycle(){
root=rebuild(0,1,tot);
}
}kd;
int main(){
ll n,lastans=0;
while(cin>>n){
kd.init();
int m=10000;
while(1){
int op=read();
if(op==3) break;
else if(op==1){
int x=read()^lastans;
int y=read()^lastans;
int v=read()^lastans;
kd.insert(kd.root,0,x,y,v);
if(kd.tot==m) kd.recycle(),m+=10000;
}else{
int x_1=read()^lastans;
int y_1=read()^lastans;
int x_2=read()^lastans;
int y_2=read()^lastans;
lastans=kd.query(kd.root,x_1,y_1,x_2,y_2);
println(lastans);
}
}
}
return 0;
}顺便贴上按size平衡因子维护的版本(TLE,不打算改了)
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 2e5+11;
const int INF = 0x7fffffff;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int D;
struct Point{
int x[2],val;
bool operator < (const Point &rhs) const{
return x[D]<rhs.x[D];
}
}p[MAXN];
struct KD{
int son[MAXN][2],lx[MAXN][2],rx[MAXN][2];
int size[MAXN];
ll sum[MAXN];
int root,tot;
struct REC{
int id;
bool operator < (const REC &rhs) const{
return p[id]<p[rhs.id];
}
}rec[MAXN];int cnt;
void dfs(int o){
if(!o) return;
dfs(son[o][0]);
rec[++cnt].id=o;
dfs(son[o][1]);
}
void pu(int o){
int lc=son[o][0],rc=son[o][1];
rep(i,0,1){
lx[o][i]=rx[o][i]=p[o].x[i];
if(lc){
if(lx[lc][i]<lx[o][i]) lx[o][i]=lx[lc][i];
if(rx[lc][i]>rx[o][i]) rx[o][i]=rx[lc][i];
}
if(rc){
if(lx[rc][i]<lx[o][i]) lx[o][i]=lx[rc][i];
if(rx[rc][i]>rx[o][i]) rx[o][i]=rx[rc][i];
}
}
sum[o]=p[o].val+sum[lc]+sum[rc];
size[o]=1+size[lc]+size[rc];
}
void init(){
root=tot=D=0;
}
bool ok(int o){
int lc=son[o][0],rc=son[o][1];
int ls=size[lc]<<2;
int rs=size[rc]<<2;
int sz=size[o]<<1|1;
return ls<=sz&&rs<=sz;
}
void insert(int &o,int now,int x,int y,int v){
if(!o){
o=++tot;
p[o].x[0]=lx[o][0]=rx[o][0]=x;
p[o].x[1]=lx[o][1]=rx[o][1]=y;
p[o].val=sum[o]=v;size[o]=1;
son[o][0]=son[o][1]=0;
return;
}else if(p[o].x[0]==x&&p[o].x[1]==y){
p[o].val+=v;sum[o]+=v;
return;
}else if(now==0){
if(x<p[o].x[0]) insert(son[o][0],1,x,y,v);
else insert(son[o][1],1,x,y,v);
}else{
if(y<p[o].x[1]) insert(son[o][0],0,x,y,v);
else insert(son[o][1],0,x,y,v);
}
pu(o);
if(size[o]>1000&&!ok(o)) rebuild(o);
}
bool inside(int x_1,int y_1,int x_2,int y_2,int X1,int Y1,int X2,int Y2){
return x_1<=X1&&X2<=x_2&&y_1<=Y1&&Y2<=y_2;
}
bool outside(int x_1,int y_1,int x_2,int y_2,int X1,int Y1,int X2,int Y2){
return x_1>X2||x_2<X1||y_1>Y2||y_2<Y1;
}
ll query(int o,int x_1,int y_1,int x_2,int y_2){
if(!o) return 0;
#define RECT lx[o][0],lx[o][1],rx[o][0],rx[o][1]
#define POT p[o].x[0],p[o].x[1],p[o].x[0],p[o].x[1]
if(inside(x_1,y_1,x_2,y_2,RECT)) return sum[o];
if(outside(x_1,y_1,x_2,y_2,RECT)) return 0;
ll res=0;
if(inside(x_1,y_1,x_2,y_2,POT)) res=p[o].val;
return res+query(son[o][0],x_1,y_1,x_2,y_2)+query(son[o][1],x_1,y_1,x_2,y_2);
}
int rebuild(int now,int l,int r){
int mid=l+r>>1;
D=now; nth_element(rec+l,rec+mid,rec+r+1);
int o=rec[mid].id;
sum[o]=p[o].val; size[o]=1;
son[o][0]=son[o][1]=0;
lx[o][0]=rx[o][0]=p[o].x[0];
lx[o][1]=rx[o][1]=p[o].x[1];
if(l<mid) son[o][0]=rebuild(now^1,l,mid-1);
if(r>mid) son[o][1]=rebuild(now^1,mid+1,r);
pu(o);
return o;
}
void rebuild(int &o){
cnt=0;
dfs(o);
o=rebuild(0,1,cnt);
}
}kd;
int main(){
ll n,lastans=0;
while(cin>>n){
kd.init();
while(1){
int op=read();
if(op==3) break;
else if(op==1){
int x=read()^lastans;
int y=read()^lastans;
int v=read()^lastans;
kd.insert(kd.root,0,x,y,v);
}else{
int x_1=read()^lastans;
int y_1=read()^lastans;
int x_2=read()^lastans;
int y_2=read()^lastans;
lastans=kd.query(kd.root,x_1,y_1,x_2,y_2);
println(lastans);
}
}
}
return 0;
}