제출 #295423

#제출 시각아이디문제언어결과실행 시간메모리
295423humbertoyustaRectangles (IOI19_rect)C++14
50 / 100
2134 ms605648 KiB
#include "rect.h"
#include<bits/stdc++.h>
#pragma GCC optimize ("Ofast")
using namespace std;
//#define int long long
#define maxn 2505
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define ll long long
#define ii pair<int,int>

int vx[202][202][202], vy[202][202][202];
map<pair<ii,ii>,bool> mp;
ll ans;
int valid[maxn][maxn], valend[maxn][maxn], mnx, mny, mxx, mxy, cnt;
bool valbgn[maxn][maxn], mk[maxn][maxn];
int n, m, A[maxn][maxn];

void dfs(int i,int j){
    mnx = min( mnx , i );
    mxx = max( mxx , i );
    mny = min( mny , j );
    mxy = max( mxy , j );
    cnt++;
    mk[i][j] = 1;
    if( i + 1 <= n - 1 && A[i+1][j] == 0 && !mk[i+1][j] ) dfs(i+1,j);
    if( i - 1 >= 0 && A[i-1][j] == 0 && !mk[i-1][j] ) dfs(i-1,j);
    if( j + 1 <= m - 1 && A[i][j+1] == 0 && !mk[i][j+1] ) dfs(i,j+1);
    if( j - 1 >= 0 && A[i][j-1] == 0 && !mk[i][j-1] ) dfs(i,j-1);
}

ll f(int x){
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            valid[i][j] = valend[i][j] = valbgn[i][j] = 0;
    for(int i=1; i<n-1; i++){
        int mx = A[i][x];
        valend[i][x-1] = 1 + i;
        valbgn[i][x-1] = 1;
        for(int j=x; j<m-1; j++){
            if( A[i][j] > mx ) mx = A[i][j];
            valid[i][j] = ( mx < A[i][x-1] && mx < A[i][j+1] );
            valend[i][j] = ( ( ( valend[i][j-1] - valend[i-1][j-1] ) != 0 ) && A[i][j] < A[i-1][j] );
            valbgn[i][j] = ( valbgn[i][j-1] && A[i][j] < A[i+1][j] );

            if( valid[i][j] ) valid[i][j] += valid[i-1][j];
            valend[i][j] += valend[i-1][j];
        }
    }
    ll res = 0;
    for(int j=x; j<m-1; j++){
        for(int i=1; i<n-1; i++){
            if( valbgn[i][j] ){
                int k = i - valid[i][j];
                res += valend[i][j] - valend[k][j];
                //if( valend[i][j] - valend[k][j] ){
                //    db(x)db(j)
                //    db(i)db(k)
                //    db(valend[i][j] - valend[k][j])
                //}
            }
        }
    }
    return res;
}

long long count_rectangles(std::vector<std::vector<int> > a) {
	n = a.size();
	m = a[0].size();
	int maximum = 0;
	for(int i=0; i<n; i++)
        for(int j=0; j<m; j++){
            A[i][j] = a[i][j];
            maximum = max( maximum , A[i][j] );
        }
    if( maximum <= 1 ){
        for(int i=1; i<n-1; i++)
            for(int j=1; j<m-1; j++)
                if( !mk[i][j] && A[i][j] == 0 ){
                    mnx = i;
                    mxx = i;
                    mny = j;
                    mxy = j;
                    cnt = 0;
                    dfs(i,j);
                    if( ( mxx - mnx + 1 ) * ( mxy - mny + 1 ) == cnt ){
                        if( mnx >= 1 && mxx <= n - 2 && mny >= 1 && mxy <= m - 2 ){
                            ans ++;
                        }
                    }
                }
    }
    else{
        if( n <= 200 && m <= 200 ){

            for(int i=1; i<n-1; i++){
                for(int j=1; j<m-1; j++){
                    int mx = A[i][j];
                    for(int k=j; k<m-1; k++){
                        if( A[i][k] > mx ) mx = A[i][k];
                        vx[i][j][k] = ( mx < A[i][j-1] && mx < A[i][k+1] );
                        if( vx[i][j][k] ){ vx[i][j][k] = vx[i][j][k] + vx[i-1][j][k]; }
                    }
                }
            }

            for(int i=1; i<m-1; i++){
                for(int j=1; j<n-1; j++){
                    int mx = A[j][i];
                    for(int k=j; k<n-1; k++){
                        if( A[k][i] > mx ) mx = A[k][i];
                        vy[i][j][k] = ( mx < A[j-1][i] && mx < A[k+1][i] );
                        if( vy[i][j][k] )
                            vy[i][j][k] += vy[i-1][j][k];
                    }
                }
            }

            for(int i=1; i<n-1; i++){
                for(int j=i; j<n-1; j++){
                    for(int k=1; k<m-1; k++){
                        for(int l=k; l<m-1; l++){
                            if( j - vx[j][k][l] < i && l - vy[l][i][j] < k ){
                                ans ++;
                            }
                        }
                    }
                }
            }

        }
        else{
            for(int i=1; i<m-1; i++)
                ans += f(i);
        }
	}
	return ans;
}
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