cf1284E. Antenna Coverage

Link

Link

Solution

就離散化一下

然後枚舉第一個會場上的位置，然後對於每個位置$x$，找出當前覆蓋$x$的集合，假設有$num$個

也就是說這$num$個區間在第一個軸上兩兩相交

那麼必須這些區間在另一個軸上也兩兩相交，那我就用線段樹維護下每個位置的覆蓋次數，必須等於$num$纔行

Code

#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 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#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
{
    ll mn[maxn<<2], mx[maxn<<2], sum[maxn<<2], add[maxn<<2], set[maxn<<2], L[maxn<<2], R[maxn<<2];
    void maketag_set(ll o, ll v)
    {
        add[o]=0;
        set[o]=v;
        mx[o]=mn[o]=v;
        sum[o]=(R[o]-L[o]+1)*v;
    }
    void maketag_add(ll o, ll v)
    {
        add[o]+=v;
        mx[o]+=v, mn[o]+=v;
        sum[o]+=(R[o]-L[o]+1)*v;
    }
    void pushdown(ll 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(ll 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(ll o, ll l, ll r, ll* array=NULL)
    {
        ll mid(l+r>>1);
        L[o]=l, R[o]=r;
        add[o]=0;
        set[o]=-1;
        if(l==r)
        {
            if(array)mn[o]=mx[o]=sum[o]=array[l];
            else mn[o]=mx[o]=sum[o]=0;
            return;
        }
        build(o<<1,l,mid,array);
        build(o<<1|1,mid+1,r,array);
        pushup(o);
    }
    void Set(ll o, ll l, ll r, ll v)
    {
        ll mid(L[o]+R[o]>>1);
        if(l<=L[o] and r>=R[o]){maketag_set(o,v);return;}
        pushdown(o);
        if(l<=mid)Set(o<<1,l,r,v);
        if(r>mid)Set(o<<1|1,l,r,v);
        pushup(o);
    }
    void Add(ll o, ll l, ll r, ll v)
    {
        ll mid(L[o]+R[o]>>1);
        if(l<=L[o] and r>=R[o]){maketag_add(o,v);return;}
        pushdown(o);
        if(l<=mid)Add(o<<1,l,r,v);
        if(r>mid)Add(o<<1|1,l,r,v);
        pushup(o);
    }
    ll Sum(ll o, ll l, ll r)
    {
        pushdown(o);
        ll mid(L[o]+R[o]>>1), ans(0);
        if(l<=L[o] and r>=R[o])return sum[o];
        if(l<=mid)ans+=Sum(o<<1,l,r);
        if(r>mid)ans+=Sum(o<<1|1,l,r);
        return ans;
    }
    ll Min(ll o, ll l, ll r)
    {
        ll mid(L[o]+R[o]>>1), ans(linf);
        if(l<=L[o] and r>=R[o])return mn[o];
        pushdown(o);
        if(l<=mid)ans=min(ans,Min(o<<1,l,r));
        if(r>mid)ans=min(ans,Min(o<<1|1,l,r));
        return ans;
    }
    ll Max(ll o, ll l, ll r)
    {
        ll mid(L[o]+R[o]>>1), ans(-linf);
        if(l<=L[o] and r>=R[o])return mx[o];
        pushdown(o);
        if(l<=mid)ans=max(ans,Max(o<<1,l,r));
        if(r>mid)ans=max(ans,Max(o<<1|1,l,r));
        return ans;
    }
}segtree;
struct Lisan
{
    int tmp[maxn], tot;
    void clear(){tot=0;}
    void insert(int x){tmp[++tot]=x;}
    void run()
    {
        sort(tmp+1,tmp+tot+1);
        tot=unique(tmp+1,tmp+tot+1)-tmp-1;
    }
    void lisan(int *a, int len)
    {
        for(int i=1;i<=len;i++)a[i]=lower_bound(tmp+1,tmp+tot+1,a[i])-tmp;
    }
    int lisan(int x)
    {
        return lower_bound(tmp+1,tmp+tot+1,x)-tmp;
    }
}ls;
vector<pll> event[maxn];
ll sa[maxn], ea[maxn], sb[maxn], eb[maxn], n;
int main()
{
    ll i, j;
    n=read();
    rep(i,1,n)
    {
        sa[i]=read(), ea[i]=read();
        sb[i]=read(), eb[i]=read();

        ls.insert(sa[i]);
        ls.insert(ea[i]);
        ls.insert(ea[i]+1);

        ls.insert(sb[i]);
        ls.insert(eb[i]);
        ls.insert(eb[i]+1);
    }
    ls.run();

    segtree.build(1,1,ls.tot);
    rep(i,1,ls.tot)event[i].clear();
    rep(i,1,n)
    {
        event[ls.lisan(sa[i])].emb(pll(i,+1));
        event[ls.lisan(ea[i]+1)].emb(pll(i,-1));
    }
    ll now=0;
    rep(i,1,ls.tot)
    {
        for(auto pr:event[i])
        {
            ll num = pr.first;
            now += pr.second;
            segtree.Add(1,ls.lisan(sb[num]),ls.lisan(eb[num]),pr.second);
        }
        if(now != segtree.Max(1,1,ls.tot))
        {
            printf("NO");
            return 0;
        }
    }

    segtree.build(1,1,ls.tot);
    rep(i,1,ls.tot)event[i].clear();
    rep(i,1,n)
    {
        event[ls.lisan(sb[i])].emb(pll(i,+1));
        event[ls.lisan(eb[i]+1)].emb(pll(i,-1));
    }
    now=0;
    rep(i,1,ls.tot)
    {
        for(auto pr:event[i])
        {
            ll num = pr.first;
            now += pr.second;
            segtree.Add(1,ls.lisan(sa[num]),ls.lisan(ea[num]),pr.second);
        }
        if(now != segtree.Max(1,1,ls.tot))
        {
            printf("NO");
            return 0;
        }
    }

    printf("YES");
    return 0;
}