[线段树] Codeforces Round #419 (Div. 1) D. Karen and Cards

从小到大枚举x，剩下的限制是y>bi⋀z>ci或者y>bi⋁z>ci 对应的是平面上一个矩形或者挖掉一个矩形
发现平面上矩形的交取个min就好了，挖掉的矩形要取并，这个可以用平衡树或线段树维护
大概是这样，我是灵魂画师

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=500005;int mx[N<<2],mn[N<<2],F[N<<2]; ll sum[N<<2];inline void mark(int x,int l,int r,int a){  mx[x]=mn[x]=a,sum[x]=(ll)(r-l+1)*a,F[x]=a;}inline void push(int x,int l,int r){  int mid=(l+r)>>1;  if (F[x]) mark(x<<1,l,mid,F[x]),mark(x<<1|1,mid+1,r,F[x]),F[x]=0;}inline void upd(int x){  mx[x]=max(mx[x<<1],mx[x<<1|1]);  mn[x]=min(mn[x<<1],mn[x<<1|1]);  sum[x]=sum[x<<1]+sum[x<<1|1];}inline void Modify(int x,int l,int r,int ql,int qr,int a){  if (ql<=l && r<=qr){    mark(x,l,r,a); return;  }  push(x,l,r); int mid=(l+r)>>1;  if (ql<=mid) Modify(x<<1,l,mid,ql,qr,a);  if (qr>mid) Modify(x<<1|1,mid+1,r,ql,qr,a);  upd(x);}inline int QL(int x,int l,int r,int a){ // <a  if (l==r) return l;  push(x,l,r); int mid=(l+r)>>1;  if (mn[x<<1]<a) return QL(x<<1,l,mid,a);  else return QL(x<<1|1,mid+1,r,a);}inline int QR(int x,int l,int r,int a){ // >=a  if (l==r) return l;  push(x,l,r); int mid=(l+r)>>1;  if (mx[x<<1|1]<a) return QR(x<<1,l,mid,a);  else return QR(x<<1|1,mid+1,r,a);}inline ll Sum(int x,int l,int r,int ql,int qr){  if (ql<=l && r<=qr) return sum[x];  push(x,l,r); int mid=(l+r)>>1; ll ret=0;  if (ql<=mid) ret+=Sum(x<<1,l,mid,ql,qr);  if (qr>mid) ret+=Sum(x<<1|1,mid+1,r,ql,qr);  return ret;}int n,p,q,r;int a[N],b[N],c[N];int idx[N];bool cmp(int x,int y){  return a[x]<a[y];}inline void Add(int x,int y){  if (mn[1]>=y) return;  int l=QL(1,1,q,y);  if (l<=x)    Modify(1,1,q,l,x,y);}inline ll Query(int x,int y){  if (x>q || y>r) return 0;   ll ret=(ll)(q-x+1)*(r-y+1);  if (mx[1]<y) return ret;  int r=QR(1,1,q,y);  if (x<=r){    ll tmp=Sum(1,1,q,x,r)-(ll)(r-x+1)*(y-1);    ret-=tmp;  }  return ret;}int maxx[N],maxy[N];int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(p); read(q); read(r);  for (int i=1;i<=n;i++) read(a[i]),read(b[i]),read(c[i]),idx[i]=i;  sort(idx+1,idx+n+1,cmp);  maxx[n+1]=1,maxy[n+1]=1;  for (int i=n;i;i--)    maxx[i]=max(maxx[i+1],b[idx[i]]+1),maxy[i]=max(maxy[i+1],c[idx[i]]+1);  int pnt=1; ll ans=0;  for (int i=1;i<=p;i++){    while (pnt<=n && a[idx[pnt]]<i) Add(b[idx[pnt]],c[idx[pnt]]),pnt++;    ans+=Query(maxx[pnt],maxy[pnt]);  }  printf("%lld\n",ans);  return 0;}