제출 #296175

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
296175		Kerim	Rectangles (IOI19_rect)	C++17	37 / 100	5102 ms	457608 KiB

rect

#include "rect.h"
#include "bits/stdc++.h"
#define MAXN 100009
#define INF 1000000007
#define mp(x,y) make_pair(x,y)
#define all(v) v.begin(),v.end()
#define pb(x) push_back(x)
#define wr cout<<"----------------"<<endl;
#define ppb() pop_back()
#define tr(ii,c) for(__typeof((c).begin()) ii=(c).begin();ii!=(c).end();ii++)
#define ff first
#define ss second
#define my_little_dodge 46
#define debug(x,y,z)  cerr<< #x <<" = "<< x<<" "<< #y <<" = "<< y<<" "<< #z <<" = "<< z<<endl;
using namespace std;

typedef long long ll;
typedef pair<int,int> PII;
template<class T>bool umin(T& a,T b){if(a>b){a=b;return 1;}return 0;}
template<class T>bool umax(T& a,T b){if(a<b){a=b;return 1;}return 0;}
const int N=2505;
int ata[N],mn[N],mx[N],arr[N],c[N],sz,sub[N],vis[N],rc,pos;
vector<int>row[N][N],col[N][N];
bool cmp(int x,int y){
	return (arr[x]<arr[y]);	
}
int tap(int x){
	if(ata[x]==x)
		return x;
	return ata[x]=tap(ata[x]);	
}
void merge(int x,int y){
	if((x=tap(x))==(y=tap(y)))
		return;
	if(sub[x]<sub[y])
		swap(x,y);
	ata[y]=x;
	sub[x]+=sub[y];
	umin(mn[x],mn[y]);
	umax(mx[x],mx[y]);	
}
map<pair<int,PII>,int>pm[2];
//The key idea is there are at most 2*n^2 such pairs
void add(int x,int y){
	assert(x+1<y);
	//if rc=0, then rectangle on row (pos) in range (arr[pos][x],arr[pos][y]) is good
	//if rc=1, then rectangle on column (pos) in range (arr[x][pos],arr[y][pos]) is good
	if(pm[rc][mp(pos,mp(x,y))])
		return;
	pm[rc][mp(pos,mp(x,y))]=1;	
	if(!rc)
		row[x][y].pb(pos);
	else
		col[x][y].pb(pos);
}
void solve(int x){
	if(x>1 and vis[x-1]){
		int a=mn[tap(x-1)]-1;
		if(a>=1)
			add(a,x);
	}	
	if(x<sz and vis[x+1]){
		int b=mx[tap(x+1)]+1;
		if(b<=sz)
			add(x,b);
	}
}
void upd(int x){
	vis[x]=1;
	if(x>1 and vis[x-1])
		merge(x-1,x);
	if(x<sz and vis[x+1])
		merge(x,x+1);	
}
void f(){
	for(int i=1;i<=sz;i++)
    	ata[i]=mn[i]=mx[i]=i,sub[i]=1,c[i]=i,vis[i]=0;
    sort(c+1,c+sz+1,cmp);
    for(int i=1;i<=sz;i++){
    	int j=i;
		while(j<=sz and arr[c[j]]==arr[c[i]])
			solve(c[j++]);
		for(int k=i;k<j;k++)
			upd(c[k]);
		i=j-1;
    }
}
long long count_rectangles(std::vector<std::vector<int> > a) {
	int n=int(a.size()),m=int(a[0].size());
	if(n<3 or m<3)
		return 0;
	{
		//precalc
		sz=m;
		for(int j=1;j<=n;j++){
			for(int i=1;i<=m;i++)
				arr[i]=a[j-1][i-1];
			rc=0;pos=j;f();
		}sz=n;
		for(int j=1;j<=m;j++){
			for(int i=1;i<=n;i++)
				arr[i]=a[i-1][j-1];
			rc=1;pos=j;f();
		}
	}
	vector<pair<PII,PII> >R,C;
	vector<int>cur;int tmp;
	for(int i=1;i<=m;i++)	
		for(int j=i+2;j<=m;j++){
			tmp=int(row[i][j].size());
			for(int k=0;k<tmp;k++){
				cur.pb(row[i][j][k]);
				if(k+1==tmp or row[i][j][k]+1!=row[i][j][k+1]){
					R.pb(mp(mp(cur[0]-1,cur.back()+1),mp(i,j)));//a,b,x,y
					cur.clear();	
				}
			}
		}
	/*
	a,b,x,y (n^2) => R
	c,d,i,j (n^2) => C
	a<=i<=j<=b
	c<=x<=y<=d
	*/
	for(int i=1;i<=n;i++)	
		for(int j=i+2;j<=n;j++){
			tmp=int(col[i][j].size());
			for(int k=0;k<tmp;k++){
				cur.pb(col[i][j][k]);
				if(k+1==tmp or col[i][j][k]+1!=col[i][j][k+1]){
					C.pb(mp(mp(cur[0]-1,cur.back()+1),mp(i,j)));//c,d,i,j
					cur.clear();	
				}
			}
		}
	ll ans=0;
	tr(r,R)
		tr(c,C)
			ans+=(r->ff.ff<=c->ss.ff and c->ss.ss<=r->ff.ss
			  and c->ff.ff<=r->ss.ff and r->ss.ss<=c->ff.ss);
	return ans;
}

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 12 / 12
• Subtask 4: 0 / 22
• Subtask 5: 10 / 10
• Subtask 6: 0 / 13
• Subtask 7: 0 / 28