bzoj 4066: 简单题 K-D树

题目大意:

http://www.lydsy.com/JudgeOnline/problem.php?id=4066

题解

我们把每次的修改操作都当作二维平面上多了一个权值点
对于每组询问可以看做求一个矩形区域内的点权和
所以我们用k-D Tree直接搜就好了

#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(int &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
const int maxn = 160010;
const int lim_siz = 2000;
int split[maxn],now;
int dem = 2;
struct Node{
	int pos[2],val;
	int minn[2],maxx[2],sum;
	Node *ch[2];
	void update(){
		for(int d=0;d<dem;++d) minn[d] = min(pos[d],min(ch[0]->minn[d],ch[1]->minn[d]));
		for(int d=0;d<dem;++d) maxx[d] = max(pos[d],max(ch[0]->maxx[d],ch[1]->maxx[d]));
		sum = val + ch[0]->sum + ch[1]->sum;
	}
 
}*null,*root,*op;
Node T[maxn];
inline bool cmp(const Node &a,const Node &b){
	return a.pos[split[now]] < b.pos[split[now]];
}
inline void init(){
	null = &T[0];
	null->ch[0] = null->ch[1] = null;
	null->val = 0;
	for(int d=0;d<dem;++d){
		null->pos[d] = 0;
		null->minn[d] = 0x3f3f3f3f;
		null->maxx[d] = -0x3f3f3f3f;
	}root = null;
}
Node* build(int l,int r,int s){
	if(l > r) return null;
	int mid = (l+r)>> 1;
	split[now = mid] = s % dem;
	nth_element(T+l,T+mid,T+r+1,cmp);
	Node *p = &T[mid];
	p->ch[0] = build(l,mid-1,s+1);
	p->ch[1] = build(mid+1,r,s+1);
	p->update();return p;
}
int sav[maxn][2],sav_cnt,cnt;
int val[maxn],cmd,X1,Y1,X2,Y2,x;
int query(Node *p){
	if(p == null) return 0;
	if(p->minn[0] >= X1 && p->minn[1] >= Y1
	&& p->maxx[0] <= X2 && p->maxx[1] <= Y2)
		return p->sum;
	else if(p->maxx[1] < Y1 || p->maxx[0] < X1 
		 || p->minn[0] > X2 || p->minn[1] > Y2)
		return 0;
	if( p->pos[0] >= X1 && p->pos[1] >= Y1
	&&  p->pos[0] <= X2 && p->pos[1] <= Y2)
		return p->val + query(p->ch[0]) + query(p->ch[1]);
	return query(p->ch[0]) + query(p->ch[1]);
}
int main(){
	init();
	int n;read(n);//read(n);
	int lastans = 0;
	while(1){
		read(cmd);
		if(cmd == 3) break;
		if(cmd == 1){
			read(X1);read(Y1);read(x);
			X1 ^= lastans;Y1 ^= lastans;x ^= lastans;
			sav[++sav_cnt][0] = X1;
			sav[sav_cnt][1] = Y1;
			val[sav_cnt] = x;
			if(sav_cnt == lim_siz){
				for(int i=1;i<=sav_cnt;++i){
					T[cnt+i].minn[0] = T[cnt+i].maxx[0] = T[cnt+i].pos[0] = sav[i][0];
					T[cnt+i].minn[1] = T[cnt+i].maxx[1] = T[cnt+i].pos[1] = sav[i][1];
					T[cnt+i].val = val[i];
				}root = build(1,cnt+=sav_cnt,1);
				sav_cnt = 0;
			}
		}else{
			read(X1);read(Y1);read(X2);read(Y2);
			X1 ^= lastans;Y1 ^= lastans;
			X2 ^= lastans;Y2 ^= lastans;
			int ans = 0;
			for(int i=1;i<=sav_cnt;++i){
				if(sav[i][0] >= X1 && sav[i][1] >= Y1
				&& sav[i][0] <= X2 && sav[i][1] <= Y2)
					ans += val[i];
			}
			ans += query(root);
			printf("%d\n",lastans = ans);
		}
	}
	getchar();getchar();
	return 0;
}