题解 [NOI2019]弹跳

题目传送门

题目大意

给出 \(n\) 做城市，每座城市都有横纵坐标 \(x,y\)。现在给出 \(m\) 个限制 \(p,t,l,r,d,u\)，表示从 \(p\) 城市出发，可以花费 \(t\) 走到 \(l\le x\le r,d\le y\le u\) 的一座城市。

问，从城市 1 出发走到每座城市的最小花费是多少？

\(n\le 7\times 10^4,m\le 1.5\times 10^5\)

思路

这道题自己只做出来 \(76\) 分，剩下的感觉还是比较妙吧？

首先不难看出，我们可以用 K-D Tree 优化建图，大概意思就是 K-D Tree 上的虚点连对应的实点，虚点连儿子，每个点连可以到的点的虚点，然后跑最短路算法就好了。

这样做可以看出边数是 \(m\sqrt n\) 级别的，所以时间复杂度是 \(\Theta(m\sqrt n\log n)\)，但是空间复杂度也是 \(m\sqrt n\) 级别的。

我们其实可以考虑不建边，每次在 K-D Tree 上进行松弛操作。这样空间复杂度就降到了 \(n+m\)。

\(\texttt{Code}\)

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define MAXN 140005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}

int n,m,w,h,rt,cut;

struct Vector{
	int x[2],pos;
	bool operator < (const Vector &p)const{return x[cut] < p.x[cut];}
}city[MAXN],pnt[MAXN];
int mi[MAXN][2],mx[MAXN][2],lson[MAXN],rson[MAXN];

void Mx (int &a,int b){a = max (a,b);}
void Mi (int &a,int b){a = min (a,b);}
void Pushup (int x){
	mi[x][0] = mx[x][0] = city[x].x[0],mi[x][1] = mx[x][1] = city[x].x[1];
	if (lson[x]) for (Int i = 0;i < 2;++ i) Mi (mi[x][i],mi[lson[x]][i]),Mx (mx[x][i],mx[lson[x]][i]);
	if (rson[x]) for (Int i = 0;i < 2;++ i) Mi (mi[x][i],mi[rson[x]][i]),Mx (mx[x][i],mx[rson[x]][i]);
}

int build (int l,int r,int k){
	if (l > r) return 0;
	int mid = (l + r) >> 1,now = mid;cut = k;
	nth_element (city + l,city + mid,city + r + 1);
	if (l == r) return Pushup (now),now;
	else{
		lson[now] = build (l,mid - 1,k ^ 1);
		rson[now] = build (mid + 1,r,k ^ 1);
	}
	Pushup (now);
	return now;
}

int dist[MAXN];
bool vis[MAXN];
#define PII pair<int,int>
priority_queue <PII,vector <PII>,greater <PII> > q; 

void Relax (int u,int w){
	if (dist[u] > w){
		dist[u] = w;
		if (!vis[u]) q.push (make_pair (dist[u],u));
	}
}

void SolveE (int now,int x1,int x2,int y1,int y2,int ind,int t){
	if (!now) return ;
	if (mi[now][0] > x2 || mx[now][0] < x1 || mi[now][1] > y2 || mx[now][1] < y1) return ;
	if (mi[now][0] >= x1 && mx[now][0] <= x2 && mi[now][1] >= y1 && mx[now][1] <= y2){
		Relax (now,dist[ind] + t);
		return ;
	}
	if (city[now].x[0] >= x1 && city[now].x[0] <= x2 && city[now].x[1] >= y1 && city[now].x[1] <= y2)
		Relax (city[now].pos,dist[ind] + t);
	SolveE (lson[now],x1,x2,y1,y2,ind,t);
	SolveE (rson[now],x1,x2,y1,y2,ind,t);
	return ;
} 

struct node{int L,R,U,D,T;};
vector <node> G[MAXN];

void Dijkstra (int s){
	memset (dist,0x7f,sizeof (dist));
	dist[s] = 0,vis[s] = 1;
	while (!q.empty()) q.pop ();
	q.push (make_pair (0,s));
	while (!q.empty()){
		PII now = q.top ();q.pop ();int u = now.second;vis[u] = 1;
		if (dist[u] != now.first) continue;
		if (u > n) for (node to : G[u - n]) SolveE (rt,to.L,to.R,to.D,to.U,u,to.T);
		else Relax (lson[u],dist[u]),Relax (rson[u],dist[u]),Relax (city[u].pos,dist[u]);
	}
}

signed main(){
	read (n,m,w,h);
	for (Int i = 1;i <= n;++ i) read (city[i].x[0],city[i].x[1]),city[i].pos = n + i;
	rt = build (1,n,0);for (Int i = 1;i <= m;++ i){
		int p,t,l,r,d,u;read (p,t,l,r,d,u);
		G[p].push_back (node {l,r,u,d,t});
	}
	Dijkstra (n + 1);
	for (Int i = 2;i <= n;++ i) write (dist[n + i]),putchar ('\n');
	return 0;
}