Board Game CodeForces

大意: 给定$n$张卡$(a_i,b_i,c_i,d_i)$, 初始坐标$(0,0)$.

假设当前在$(x,y)$, 若$xge a_i,yge b_i$, 则可以使用第$i$张卡, 使用后达到坐标$c_i,d_i$.

求最少使用多少张卡后才能使用最后一张卡.

直接$BFS$的话是$O(n^2)$的, 用数据结构优化即可.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <map>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
using namespace std;
typedef pair<int,int> pii;
typedef tuple<int,int,int> t3;
const int N = 1e6+10;
const int INF = 0x3f3f3f3f;
int n, cnt, b[N], pre[N];
struct {int a,b,c,d,id;} a[N];
queue<t3> q;
priority_queue<pii,vector<pii>,greater<pii> > s[N];
pii tr[N<<2];
int ID(int x) {
	return lower_bound(b+1,b+1+*b,x)-b;
}
void dfs(int x) {
	++cnt;
	if (pre[x]!=-1) dfs(pre[x]);
	else printf("%d
",cnt);
	printf("%d ",x);
}

void build(int o, int l, int r) {
	if (l==r) {
		tr[o].x = s[l].empty()?INF:s[l].top().x;
		tr[o].y = l;
		return;
	}
	build(ls),build(rs);
	tr[o]=min(tr[lc],tr[rc]);
}
void update(int o, int l, int r, int x) {
	if (l==r) {
		tr[o].x = s[l].empty()?INF:s[l].top().x;
		return;
	}
	mid>=x?update(ls,x):update(rs,x);
	tr[o]=min(tr[lc], tr[rc]);
}
pii find(int o, int l, int r, int ql, int qr) {
	if (ql<=l&&r<=qr) return tr[o];
	if (mid>=qr) return find(ls,ql,qr);
	if (mid<ql) return find(rs,ql,qr);
	return min(find(ls,ql,qr),find(rs,ql,qr));
}

int main() {
	scanf("%d", &n);
	REP(i,1,n) { 
		scanf("%d%d%d%d",&a[i].a,&a[i].b,&a[i].c,&a[i].d);
		b[++*b]=a[i].a;
		b[++*b]=a[i].b;
		b[++*b]=a[i].c;
		b[++*b]=a[i].d;
		a[i].id = i;
	}
	b[++*b] = 0;
	sort(b+1,b+1+*b),*b=unique(b+1,b+1+*b)-b-1;
	REP(i,1,n) {
		a[i].a=ID(a[i].a);
		a[i].b=ID(a[i].b);
		a[i].c=ID(a[i].c);
		a[i].d=ID(a[i].d);
		s[a[i].b].push(pii(a[i].a,a[i].id));
	}
	build(1,1,*b);
	q.push(t3(ID(0),ID(0),-1));
	while (q.size()) {
		t3 u = q.front(); q.pop();
		int x, y, id;
		tie(x,y,id) = u;
		while (1) {
			pii ret = find(1,1,*b,1,y);
			if (ret.x>x) break;
			while (s[ret.y].size()&&s[ret.y].top().x<=x) {
				int id2 = s[ret.y].top().y; s[ret.y].pop();
				pre[id2] = id;
				q.push(t3(a[id2].c,a[id2].d,id2));
			}
			update(1,1,*b,ret.y);
		}
	}
	if (!pre[n]) return puts("-1"),0;
	dfs(n),hr;
}