【题解】 「NOI2019」弹跳 KDT优化建图+最短路+剪枝+卡空间 LOJ3159

Legend

有 (n (1 le n le 70000)) 个二维平面上的点，(m (1 le m le 150000)) 条集体边，每条集体边为从某个点向一个矩形所在的所有点连边。

求 (1) 到所有点的最短路。

时空限制：( extrm{2s/128MB})。

Editorial

很容易让人想到 ( m{xxx}) 优化建图，发现非常无趣的是，二维线段树是两个 (log)。被卡啦！

于是我们果断用 ( m{kdtree})！发现还是被卡空间啦！

于是我们考虑不把边建出来，而是直接对整棵子树取 (min)。

( m{kdtree}) 恰好是一棵平衡树，也可以维护全局最小值！

所以你就可以很方便的写 ( m{Dijkstra}) 啦！

复杂度？(O(m sqrt{n}))！

Code

• ( m{kdtree}) 真的难写。

• ( m{kdtree}) 不剪枝？( m{TLE}) 欢迎你。

#include <bits/stdc++.h>

#define debug(...) ;
// fprintf(stderr ,__VA_ARGS__)
#define __FILE(x)
	freopen(#x".in" ,"r" ,stdin);
	freopen(#x".out" ,"w" ,stdout)

int read(){
	char k = getchar(); int x = 0;
	while(k < '0' || k > '9') k = getchar();
	while(k >= '0' && k <= '9') x = x * 10 + k - '0' ,k = getchar();
	return x;
}

inline void chkmin(int &x ,int y){x = std::min(x ,y);}
inline void chkmax(int &x ,int y){x = std::max(x ,y);}

const int MX = 2e5 + 233;
struct point{
	int x[2] ,w ,id;
	point(int X ,int Y ,int W = 0 ,int ID = 0){x[0] = X ,x[1] = Y ,w = W ,id = ID;}
	point(int a[2] ,int W = 0 ,int ID = 0){x[0] = a[0] ,x[1] = a[1] ,w = W ,id = ID;}
	point(){x[0] = x[1] = id = 0 ,w = INT_MAX;}
}P[MX];

struct node{
	int mn[2] ,mx[2] ,ch[2];
	int del ,covmn ,mnval ,mnfr ,val ,alive;
	int mxval;
	point tp;
};

int D;
bool operator <(point a ,point b){return a.x[D] < b.x[D];}

int trans[MX] ,tkdt[MX];
struct KDT{
	#define lch tr[x].ch[0]
	#define rch tr[x].ch[1]
	node tr[MX] ;
	int cnt ,rt ,D;
	KDT(){
		for(int i : {0 ,1}){
			tr[0].mx[i] = INT_MIN;
			tr[0].mn[i] = INT_MAX;
		}
		tr[0].covmn = tr[0].mnval = tr[0].val = INT_MAX;
		tr[0].mnfr = tr[0].alive = false;
		tr[0].del = true;
		tr[0].mxval = INT_MIN;
	}
	void docovmn(int x ,int v){
		if(!tr[x].del) chkmin(tr[x].val ,v);
		if(tr[x].alive){
			chkmin(tr[x].mnval ,v);
			chkmin(tr[x].covmn ,v);
			chkmin(tr[x].mxval ,v);
		}
	}
	void _pushup(int x){

		if(!tr[x].del){
			tr[x].mnval = tr[x].val;
			tr[x].mxval = tr[x].val;
			tr[x].mnfr = x;
		}
		else{
			tr[x].mnfr = 0 ,tr[x].mnval = INT_MAX;
			tr[x].mxval = INT_MIN;
		}

		tr[x].alive = tr[lch].alive || tr[rch].alive || !tr[x].del;

		for(int i : {0 ,1}){
			tr[x].mn[i] = tr[x].mx[i] = tr[x].tp.x[i];
			for(int j : {lch ,rch}){
				chkmin(tr[x].mn[i] ,tr[j].mn[i]);
				chkmax(tr[x].mx[i] ,tr[j].mx[i]);
				chkmin(tr[x].mnval ,tr[j].mnval);
				chkmax(tr[x].mxval ,tr[j].mxval);
				if(tr[x].mnval == tr[j].mnval){
					tr[x].mnfr = tr[j].mnfr;
				}
			}
		}
	}
	void pushup(int x){

		if(!tr[x].del){
			tr[x].mnval = tr[x].val;
			tr[x].mxval = tr[x].val;
			tr[x].mnfr = x;
		}
		else{
			tr[x].mnfr = 0 ,tr[x].mnval = INT_MAX;
			tr[x].mxval = INT_MIN;
		}
		tr[x].alive = tr[lch].alive || tr[rch].alive || !tr[x].del;

		for(int j : {lch ,rch}){
			chkmin(tr[x].mnval ,tr[j].mnval);
			chkmax(tr[x].mxval ,tr[j].mxval);
			if(tr[x].mnval == tr[j].mnval){
				tr[x].mnfr = tr[j].mnfr;
			}
		}
	}
	void pushdown(int x){
		if(tr[x].covmn != INT_MAX){
			if(lch) docovmn(lch ,tr[x].covmn);
			if(rch) docovmn(rch ,tr[x].covmn);
			tr[x].covmn = INT_MAX;
		}
	}
	void outputinfo(int x){
		debug("Info of vertex %d:
" ,x);
		debug("> val = %d ,mnval = %d ,mnvalfrom = %d
" ,tr[x].val ,tr[x].mnval ,tr[x].mnfr);
		debug("> point = (%d ,%d)
" ,tr[x].tp.x[0] ,tr[x].tp.x[1]);
	}
	int build(int L ,int R ,int d = 1){
		if(L > R) return 0;
		int mid = (L + R) >> 1 ,x = mid;

		D = d ,std::nth_element(P + L ,P + mid ,P + R + 1);
		trans[mid] = P[mid].id;
		tkdt[P[mid].id] = mid;
		
		tr[x].tp = P[mid];
		tr[x].covmn = INT_MAX;
		tr[x].val = tr[x].tp.w;

		lch = build(L ,mid - 1 ,!d);
		rch = build(mid + 1 ,R ,!d);
		_pushup(x);
		// outputinfo(x);
		return x;
	}
	void buildtree(int L ,int R){rt = build(L ,R);}
	bool inside(point A ,point B ,point C ,point D){
		// check if AB is in CD
		for(int i : {0 ,1}){
			if(!(C.x[i] <= A.x[i] && B.x[i] <= D.x[i])){
				return false;
			}
		}
		return true;
	}
	bool outside(point A ,point B ,point C ,point D){
		return B.x[0] < C.x[0] || A.x[0] > D.x[0] 
			|| B.x[1] < C.x[1] || A.x[1] > D.x[1];
	}
	void upd(point A ,point B ,int x ,int v){
		if(!x) return;
		pushdown(x);
		if(inside(tr[x].mn ,tr[x].mx ,A ,B)){return docovmn(x ,v);}
		if(tr[x].mxval <= v || outside(tr[x].mn ,tr[x].mx ,A ,B) || !tr[x].alive){return;}
		if(inside(tr[x].tp ,tr[x].tp ,A ,B) && !tr[x].del){
			debug("upd (%d ,%d) by %d
" ,tr[x].tp.x[0] ,tr[x].tp.x[1] ,v);
			tr[x].val = std::min(tr[x].val ,v);
		}
		upd(A ,B ,lch ,v) ,upd(A ,B ,rch ,v);
		return pushup(x);
	}
	void del(point A ,int x){
		if(!x) return ;
		if(outside(A ,A ,tr[x].mn ,tr[x].mx) || !tr[x].alive) return;
		pushdown(x);
		if(inside(tr[x].tp ,tr[x].tp ,A ,A)){
			tr[x].del = true;
			pushup(x);
			return ;
		}
		del(A ,lch) ,del(A ,rch);
		pushup(x);
	}
	std::pair<int ,int> getmn(){
		return std::make_pair(tr[rt].mnval ,tr[rt].mnfr);
	}
	void pushall(int x){
		if(!x) return;
		pushdown(x);
		pushall(lch) ,pushall(rch);
		pushup(x);
		outputinfo(x);
	}
}kdt;

struct Edge{
	int w;
	point A ,B;
	Edge(){w = 0 ,A = B = point();}
	Edge(int dist ,point a ,point b){
		w = dist;
		A = a ,B = b;
	}
};
std::vector<Edge> e[MX];
int dis[MX];

int main(){
	__FILE([NOI2019]弹跳);
	int n = read() ,m = read() ,w = read() ,h = read();
	for(int i = 1 ,x ,y ; i <= n ; ++i){
		x = read() ,y = read();
		P[i] = point(x ,y ,(INT_MAX) >> 1 ,i);
	}
	kdt.buildtree(1 ,n);
	for(int i = 1 ,p ,t ,L1 ,R1 ,L2 ,R2 ; i <= m ; ++i){
		p = read() ,t = read();
		L1 = read() ,L2 = read() ,R1 = read() ,R2 = read();
		e[p].push_back(Edge(t ,point(L1 ,R1) ,point(L2 ,R2)));
	}
	memset(dis ,0x3f ,sizeof dis);
	kdt.upd(P[tkdt[1]] ,P[tkdt[1]] ,kdt.rt ,0);
	for(int i = 1 ; i <= n ; ++i){
		// kdt.pushall(kdt.rt);
		std::pair<int ,int> cur = kdt.getmn();
		int x = trans[cur.second] ,dist = cur.first;
		debug("%d updated
" ,x);
		kdt.del(P[cur.second] ,kdt.rt);
		dis[x] = dist;
		for(auto j : e[x]){
			debug("$ TRY to update from %d...
" ,x);
			kdt.upd(j.A ,j.B ,kdt.rt ,dist + j.w);
		}
	}
	for(int i = 2 ; i <= n ; ++i){
		printf("%d
" ,dis[i]);
	}
	return 0;
}