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题解

其实就是查询一个子矩形中是否每一行都有一个車，或者每列都有一个車，这两个条件满足其一就是$yes$

现在考虑怎么查询是否每行都有一个車

那么其实就是把$[x_1,x_2]$这几行的車拿出来，然后看看纵坐标在$y_1,y_2$的那些，是不是在$[x_1,x_2]$每行都出现了，我把所有的車按照纵坐标排序，从左往右依次加入，当我把$y_2$这一排的都加入完了之后，我只需要看下$[x_1,x_2]$这些行中每行最后一个出现的車纵坐标是否大于等于$y_1$就行了

代码

#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> #define iinf 0x3f3f3f3f #define linf (1ll<<60) #define eps 1e-8 #define maxn 200010 #define maxe 100010 #define cl(x) memset(x,0,sizeof(x)) #define rep(_,__) for(_=1;_<=(__);_++) #define em(x) emplace(x) #define emb(x) emplace_back(x) #define emf(x) emplace_front(x) #define fi first #define se second #define de(x) cerr<<#x<<" = "<<x<<endl using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef pair<int,int> pii; typedef pair<ll,ll> pll; ll read(ll x=0) { ll c, f(1); for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f; for(;isdigit(c);c=getchar())x=x*10+c-0x30; return f*x; } struct SegmentTree { #define inf 0x3f3f3f3f int mn[maxn<<2], mx[maxn<<2], sum[maxn<<2], add[maxn<<2], set[maxn<<2], L[maxn<<2], R[maxn<<2]; void maketag_set(int o, int v) { add[o]=0; set[o]=v; mx[o]=mn[o]=v; sum[o]=(R[o]-L[o]+1)*v; } void maketag_add(int o, int v) { add[o]+=v; mx[o]+=v, mn[o]+=v; sum[o]+=(R[o]-L[o]+1)*v; } void pushdown(int o) { if(L[o]==R[o])return; if(~set[o]) { maketag_set(o<<1,set[o]); maketag_set(o<<1|1,set[o]); set[o]=-1; } if(add[o]) { maketag_add(o<<1,add[o]); maketag_add(o<<1|1,add[o]); add[o]=0; } } void pushup(int o) { mx[o]=max(mx[o<<1],mx[o<<1|1]); mn[o]=min(mn[o<<1],mn[o<<1|1]); sum[o]=sum[o<<1]+sum[o<<1|1]; } void build(int o, int l, int r) { int mid(l+r>>1); L[o]=l, R[o]=r; add[o]=0; set[o]=-1; mx[o]=mn[o]=sum[o]=0; if(l==r)return; build(o<<1,l,mid); build(o<<1|1,mid+1,r); pushup(o); } void segset(int o, int l, int r, int v) { int mid(L[o]+R[o]>>1); if(l<=L[o] and r>=R[o]){maketag_set(o,v);return;} pushdown(o); if(l<=mid)segset(o<<1,l,r,v); if(r>mid)segset(o<<1|1,l,r,v); pushup(o); } void segadd(int o, int l, int r, int v) { int mid(L[o]+R[o]>>1); if(l<=L[o] and r>=R[o]){maketag_add(o,v);return;} pushdown(o); if(l<=mid)segadd(o<<1,l,r,v); if(r>mid)segadd(o<<1|1,l,r,v); pushup(o); } int segsum(int o, int l, int r) { pushdown(o); int mid(L[o]+R[o]>>1), ans(0); if(l<=L[o] and r>=R[o])return sum[o]; if(l<=mid)ans+=segsum(o<<1,l,r); if(r>mid)ans+=segsum(o<<1|1,l,r); return ans; } int segmin(int o, int l, int r) { int mid(L[o]+R[o]>>1), ans(inf); if(l<=L[o] and r>=R[o])return mn[o]; pushdown(o); if(l<=mid)ans=min(ans,segmin(o<<1,l,r)); if(r>mid)ans=min(ans,segmin(o<<1|1,l,r)); return ans; } int segmax(int o, int l, int r) { int mid(L[o]+R[o]>>1), ans(-inf); if(l<=L[o] and r>=R[o])return mx[o]; pushdown(o); if(l<=mid)ans=max(ans,segmax(o<<1,l,r)); if(r>mid)ans=max(ans,segmax(o<<1|1,l,r)); return ans; } #undef inf }segtree; pll rook[maxn]; ll ans[maxn], id[maxn]; pair<pll,pll> r[maxn]; int main() { ll i, j, n, m, k, q, p; n=read(), m=read(), k=read(), q=read(); rep(i,k) rook[i].fi=read(), rook[i].se=read(); rep(i,q) r[i].fi.fi=read(), r[i].fi.se=read(), r[i].se.fi=read(), r[i].se.se=read(), id[i]=i; sort(rook+1,rook+k+1,[](pll p1, pll p2){return p1.se<p2.se;}); sort(id+1,id+q+1,[](ll a, ll b){return r[a].se.se<r[b].se.se;}); segtree.build(1,1,n); for(j=1,p=1;j<=q;j++) { while(p<=k and rook[p].se<=r[id[j]].se.se) segtree.segset(1,rook[p].fi,rook[p].fi,rook[p].se), p++; ans[id[j]]+=(segtree.segmin(1,r[id[j]].fi.fi,r[id[j]].se.fi)>=r[id[j]].fi.se); } sort(rook+1,rook+k+1,[](pll p1, pll p2){return p1.fi<p2.fi;}); sort(id+1,id+q+1,[](ll a, ll b){return r[a].se.fi<r[b].se.fi;}); segtree.build(1,1,m); for(j=1,p=1;j<=q;j++) { while(p<=k and rook[p].fi<=r[id[j]].se.fi) segtree.segset(1,rook[p].se,rook[p].se,rook[p].fi), p++; ans[id[j]]+=(segtree.segmin(1,r[id[j]].fi.se,r[id[j]].se.se)>=r[id[j]].fi.fi); } rep(i,q)if(ans[i])printf("YES\n"); else printf("NO\n"); return 0; }