UVA11297 Census

Description

給定一個$N \times N$的二維平面，你需要支持$q$個如下操作：

• $q \: x_1 \: y_1 \: x_2 \: y_2$ 找出 $(x_1, y_1) - (x_2, y_2)$ 的最大值和最小值
• $c\: x \: y \: v$ 把 $(x, y)$ 的值改爲 $v$

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Input

一行輸入$n$。

下面$n$行，每行$n$個數，表示$n \times n$矩陣。

一行輸入$q$。

下面$q$行，輸入$q$個操作。

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Output

對於每個$q$操作，一行輸出答案。

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Solution

萬惡的pdf，題面打得好累。。。

寫什麼樹套樹，KD-tree多好。

一看範圍最大/小，KD-tree能搞！
按照模板寫寫就好了。

修改直接改點權，記錄下每個點的編號，暴力改完一路上跳更新就好了。

要卡卡常。

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Code

#include<bits/stdc++.h>
using namespace std;
const double alpha=0.75;
struct node{
	int d[2],w;
	void make(int x,int y,int z){
		d[0]=x,d[1]=y,w=z;
	}
}a[1000010];
struct data{
	int mx[2],mn[2],sz,ls,rs,maxn,minn,fa;
	node p;
}tree[1000010];
int stk[1000010],mp[510][510],top,tot,word,rt,n,q,ans1,ans2;
bool operator <(node u,node v){
	return u.d[word]<v.d[word];
}
int newnode(){
	if(top) return stk[top--];
	return ++tot; 
}
void up(int x){
	int l=tree[x].ls;
	int r=tree[x].rs;
	for(int i=0;i<2;i++){
		tree[x].mn[i]=tree[x].mx[i]=tree[x].p.d[i];
		if(l) tree[x].mn[i]=min(tree[x].mn[i],tree[l].mn[i]);
		if(r) tree[x].mn[i]=min(tree[x].mn[i],tree[r].mn[i]);
		if(l) tree[x].mx[i]=max(tree[x].mx[i],tree[l].mx[i]);
		if(r) tree[x].mx[i]=max(tree[x].mx[i],tree[r].mx[i]);
	}
	tree[x].maxn=tree[x].minn=tree[x].p.w;
	mp[tree[x].p.d[0]][tree[x].p.d[1]]=x;
	if(l){
		tree[x].maxn=max(tree[x].maxn,tree[l].maxn);
		tree[x].minn=min(tree[x].minn,tree[l].minn);
	}
	if(r){
		tree[x].maxn=max(tree[x].maxn,tree[r].maxn);
		tree[x].minn=min(tree[x].minn,tree[r].minn);
	}
	tree[x].sz=tree[l].sz+tree[r].sz+1;
}
int build(int l,int r,int wd){
	if(l>r) return 0;
	int x=newnode();
	int mid=(l+r)>>1;
	word=wd,nth_element(a+l,a+mid,a+r+1);
	tree[x].p=a[mid];
	tree[x].ls=build(l,mid-1,wd^1);
	tree[x].rs=build(mid+1,r,wd^1);
	int ls=tree[x].ls,rs=tree[x].rs;
	if(ls) tree[ls].fa=x;
	if(rs) tree[rs].fa=x;
	up(x); return x;
}
void flt(int x,int num){
	int l=tree[x].ls;
	int r=tree[x].rs;
	if(l) flt(l,num);
	stk[++top]=x;
	a[tree[l].sz+num+1]=tree[x].p;
	if(r) flt(r,num+tree[l].sz+1);
}
void check(int &x,int wd){
	int l=tree[x].ls,r=tree[x].rs;
	if(tree[x].sz*alpha<tree[l].sz||tree[x].sz*alpha<tree[r].sz)
	flt(x,0),x=build(1,tree[x].sz,wd);
}
void update(int x,int v){//暴力上跳
	tree[x].p.w=v;
	while(x){
		up(x);
		x=tree[x].fa;
	}
} 
void query1(int x,int x1,int y1,int x2,int y2){
	if(x==0) return;
	if(tree[x].maxn<=ans1) return;
	int l=tree[x].ls,r=tree[x].rs;
	if(x1<=tree[x].mn[0]&&tree[x].mx[0]<=x2&&y1<=tree[x].mn[1]&&tree[x].mx[1]<=y2){
		ans1=max(ans1,tree[x].maxn);
		return;
	}
	if(x1>tree[x].mx[0]||tree[x].mn[0]>x2||y1>tree[x].mx[1]||tree[x].mn[1]>y2) return;
	if(x1<=tree[x].p.d[0]&&tree[x].p.d[0]<=x2&&y1<=tree[x].p.d[1]&&tree[x].p.d[1]<=y2)
	ans1=max(ans1,tree[x].p.w);
	if(tree[l].maxn>tree[r].maxn){
		query1(l,x1,y1,x2,y2);
		query1(r,x1,y1,x2,y2);
	}
	else{
		query1(r,x1,y1,x2,y2);
		query1(l,x1,y1,x2,y2);
	}
}
void query2(int x,int x1,int y1,int x2,int y2){
	if(x==0) return;
	if(tree[x].minn>=ans2) return;
	int l=tree[x].ls,r=tree[x].rs;
	if(x1<=tree[x].mn[0]&&tree[x].mx[0]<=x2&&y1<=tree[x].mn[1]&&tree[x].mx[1]<=y2){
		ans2=min(ans2,tree[x].minn);
		return;
	}
	if(x1>tree[x].mx[0]||tree[x].mn[0]>x2||y1>tree[x].mx[1]||tree[x].mn[1]>y2) return;
	if(x1<=tree[x].p.d[0]&&tree[x].p.d[0]<=x2&&y1<=tree[x].p.d[1]&&tree[x].p.d[1]<=y2)
	ans2=min(ans2,tree[x].p.w);
	query2(l,x1,y1,x2,y2);
	query2(r,x1,y1,x2,y2);
}
int main(){
	char s[3];
	int x1,y1,x2,y2,v;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++){
		scanf("%d",&v);
		a[(i-1)*n+j].make(i,j,v);
	}
	scanf("%d",&q);
	rt=build(1,n*n,0);
	for(int i=1;i<=q;i++){
		scanf("%s%d%d",s,&x1,&y1);
		if(s[0]=='q'){
			ans1=-2e9,ans2=2e9;
			scanf("%d%d",&x2,&y2);
			query1(rt,x1,y1,x2,y2);
			query2(rt,x1,y1,x2,y2);
			printf("%d %d\n",ans1,ans2);
		}
		else{
			scanf("%d",&v);
			update(mp[x1][y1],v); 
		}
	}
}