[Treap 笛卡尔树 扫描线 补集转化] BZOJ 2658 [Zjoi2012]小蓝的好友(mrx)

这个先转化成全部矩形 C2R+1∗C2C+1 减去不包含点的矩形
我们考虑扫描线 也就是以当前直线作为右边界的矩形有多少个
对于每个纵坐标 设h表示最左能扩展到哪里
假设hmin(1,n)=hx
那么答案至少应该C2C+1∗hmin
然后还要加上C2x−1∗(hmin(1,x−1)−hx) 和 C2C−x∗(hmin(x+1,C)−hx)
然后还要继续递归式的加 这个可以用笛卡尔树维护
扫描线向右移就h都+1 碰到点就h降为0

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef pair<int,int> abcd;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=100005;int n,R,C; abcd a[N];ll Ans;inline ll calc(int n){ return (ll)n*(n+1)/2; }struct node{  node *l,*r; int size;  int add,h; ll ans;  void upd(){    size=1,ans=0;    if (l) size+=l->size,ans+=l->ans+(ll)(l->h-h)*calc(l->size);    if (r) size+=r->size,ans+=r->ans+(ll)(r->h-h)*calc(r->size);  }  void mark(int t){ add+=t,h+=t; }  void push(){ if(l)l->mark(add); if(r)r->mark(add); add=0; }}nodes[N],*root;typedef pair<node*,node*> Droot;inline int ran(){  static int x=31253125; x+=(x<<4)+1; return x&65536;}inline node* M(node *x,node *y){  if (!x || !y) return x?x:y;  x->push(); y->push();  if (x->h>y->h || (x->h==y->h && ran()))    return y->l=M(x,y->l),y->upd(),y;  else    return x->r=M(x->r,y),x->upd(),x;}inline int Size(node *x){ return x?x->size:0; }inline Droot S(node *x,int k){  if(!x) return Droot(NULL,NULL);  x->push(); Droot y;  if (Size(x->l)>=k){    y=S(x->l,k);    x->l=y.second; x->upd();    y.second=x;  }else{    y=S(x->r,k-Size(x->l)-1);    x->r=y.first; x->upd();    y.first=x;  }  return y;}int main(){  freopen("mrx.in","r",stdin);  freopen("mrx.out","w",stdout);  read(R); read(C); read(n);  for (int i=1;i<=n;i++) read(a[i].first),read(a[i].second);  sort(a+1,a+n+1);   Ans=calc(R)*calc(C);  for (int i=1;i<=C;i++) nodes[i].size=1,root=M(root,nodes+i);  int pnt=1;  for (int i=1;i<=R;i++){    root->mark(1);    while (pnt<=n && a[pnt].first==i){      Droot t1=S(root,a[pnt].second-1),t2=S(t1.second,1);      t2.first->h=0;      root=M(M(t1.first,t2.first),t2.second);      pnt++;    }    Ans-=root->ans+(ll)root->h*calc(root->size);  }  printf("%lld\n",Ans);  return 0;}