Nanami's Digital Board CodeForces - 434B (棋盘dp)

大意: 给定01矩阵, m个操作, 操作1翻转一个点, 操作2求边界包含给定点的最大全1子矩阵

暴力枚举矩形高度, 双指针统计答案

#include <iostream>
#include <algorithm>
#include <math.h>
#include <cstdio>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <string.h>
#include <queue>
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;

const int N = 1e3+10, INF = 0x3f3f3f3f;
int n, m, q;
int a[N][N], U[N][N], D[N][N], L[N][N], R[N][N];

void init() {
	REP(i,1,n) REP(j,1,m) if (a[i][j]) {
		U[i][j]=U[i-1][j]+1;
		L[i][j]=L[i][j-1]+1;
	}
	PER(i,1,n) PER(j,1,m) if (a[i][j]) {
		D[i][j]=D[i+1][j]+1;
		R[i][j]=R[i][j+1]+1;
	}
}

void update(int x, int y) {
	a[x][y] ^= 1;
	REP(i,1,n) U[i][y]=(a[i][y]?U[i-1][y]+1:0);
	PER(i,1,n) D[i][y]=(a[i][y]?D[i+1][y]+1:0);
	REP(i,1,m) L[x][i]=(a[x][i]?L[x][i-1]+1:0);
	PER(i,1,m) R[x][i]=(a[x][i]?R[x][i+1]+1:0);
}

int query(int x, int y) {
	int ans = 0, t = U[x][y], l = y, r = y;
	PER(i,1,U[x][y]) {
		while (l>1&&U[x][l-1]>=i) --l;
		while (r<m&&U[x][r+1]>=i) ++r;
		if (U[x][l]>=i&&U[x][r]>=i) {
			ans = max(ans, i*(r-l+1));
		}
	}
	t = D[x][y], l = y, r = y;
	PER(i,1,D[x][y]) {
		while (l>1&&D[x][l-1]>=i) --l;
		while (r<m&&D[x][r+1]>=i) ++r;
		if (D[x][l]>=i&&D[x][r]>=i) {
			ans = max(ans, i*(r-l+1));
		}
	}
	t = L[x][y], l = x, r = x;
	PER(i,1,L[x][y]) {
		while (l>1&&L[l-1][y]>=i) --l;
		while (r<n&&L[r+1][y]>=i) ++r;
		if (L[l][y]>=i&&L[r][y]>=i) {
			ans = max(ans, i*(r-l+1));
		}
	}
	t = R[x][y], l = x, r = x;
	PER(i,1,R[x][y]) {
		while (l>1&&R[l-1][y]>=i) --l;
		while (r<n&&R[r+1][y]>=i) ++r;
		if (R[l][y]>=i&&R[r][y]>=i) {
			ans = max(ans, i*(r-l+1));
		}
	}
	return ans;
}

int main() {
	scanf("%d%d%d", &n, &m, &q);
	REP(i,1,n) REP(j,1,m) scanf("%d", &a[i][j]);
	init();
	REP(i,1,q) {
		int op, x, y;
		scanf("%d%d%d", &op, &x, &y);
		if (op==1) update(x,y);
		else printf("%d\n", query(x,y));
	}
}

Nanami's Digital Board CodeForces - 434B (棋盘dp)

原文地址：https://www.cnblogs.com/uid001/p/10436335.html