UOJ#191. 【集训队互测2016】Unknown

题目描述

Solution

二进制分组。
每一个组内维护一个斜率单调减的凸包。
因为有删点，避免出现反复横跳产生的爆炸复杂度，需要等到同一深度的下一个区间填满之后再合并当前区间。
时间复杂度 $O(nlg^2n)$ 。

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
struct Vct
{int x,y;Vct(int x1=0,int y1=0) { x=x1,y=y1; }Vct operator + (const Vct &a) { return Vct(x+a.x,y+a.y); }Vct operator - (const Vct &a) { return Vct(x-a.x,y-a.y); }ll operator * (const Vct &a) { return 1ll*x*a.y-1ll*y*a.x; }friend bool operator < (const Vct &a,const Vct &b) { return (a.x<b.x)||(a.x==b.x&&a.y<b.y); }
};
struct Convex_Hull
{vector<Vct> V;void Clear() { V.clear(); }Convex_Hull(){ Clear(); }void Push(Vct x) { int n=V.size(); while (n>=2&&(x-V[n-2])*(V[n-1]-V[n-2])<=0) V.pop_back(),n--; n++,V.PB(x); }friend Convex_Hull Merge(const Convex_Hull &x,const Convex_Hull &y){Convex_Hull ans;int i,j;for (i=0,j=0;i<x.V.size()&&j<y.V.size();) ans.Push(x.V[i]<y.V[j]?x.V[i++]:y.V[j++]);for (;i<x.V.size();) ans.Push(x.V[i++]);for (;j<y.V.size();) ans.Push(y.V[j++]);return ans;}ll Query(Vct x){int l=0,r=V.size()-1;while (l+5<r){int mid=(l+r+1)>>1;if ((V[mid]-V[mid-1])*x<=0) l=mid;else r=mid-1;}ll ans=-loo;for (int i=l;i<=r;i++) upmax(ans,x*V[i]);return ans;}
};
struct Segment_Tree
{int flag[MAXN<<1],lst[20];Convex_Hull tree[MAXN<<1];void up(int x) { flag[x]=1,tree[x]=Merge(tree[x<<1],tree[x<<1|1]); }void build(int x,int l,int r,int dep){lst[dep]=0;tree[x].Clear(),flag[x]=0;if (l==r) return;int mid=(l+r)>>1;build(x<<1,l,mid,dep+1);build(x<<1|1,mid+1,r,dep+1);}void insert(int x,int l,int r,int y,Vct z,int dep){if (r==y){if (lst[dep]) up(lst[dep]);lst[dep]=(l==r)?0:x;}if (l==r) { tree[x].Push(z); return; }int mid=(l+r)>>1;if (y<=mid) insert(x<<1,l,mid,y,z,dep+1);else insert(x<<1|1,mid+1,r,y,z,dep+1);}void erase(int x,int l,int r,int y,int dep){tree[x].Clear(),flag[x]=0;if (lst[dep]==x) lst[dep]=0;if (l==r) return; int mid=(l+r)>>1;if (y<=mid) erase(x<<1,l,mid,y,dep+1);else erase(x<<1|1,mid+1,r,y,dep+1);}ll query(int x,int l,int r,int L,int R,Vct y){int mid=(l+r)>>1;if (L<=l&&r<=R) {if (l==r||flag[x]) return tree[x].Query(y);return max(query(x<<1,l,mid,l,mid,y),query(x<<1|1,mid+1,r,mid+1,r,y));}if (R<=mid) return query(x<<1,l,mid,L,R,y);else if (L>mid) return query(x<<1|1,mid+1,r,L,R,y);else return max(query(x<<1,l,mid,L,mid,y),query(x<<1|1,mid+1,r,mid+1,R,y));}
} segment;
int main()
{int zbl=read(),Case;while (Case=read()){int n=min(300000,Case),num=0,ans=0;segment.build(1,1,n,0);while (Case--){int opt=read();if (opt==1) { int x=read(),y=read(); segment.insert(1,1,n,++num,Vct(x,y),0); }else if (opt==2) { segment.erase(1,1,n,num--,0); }else {int l=read(),r=read(),x=read(),y=read();ll t=segment.query(1,1,n,l,r,Vct(x,y));((t%=mods)+=mods)%=mods;ans^=t;}}printf("%d\n",ans);}return 0;
}