【YbtOJ高效进阶 哈希-3】对称正方形

链接

YbtOJ高效进阶 哈希-3

题目描述

给定一个n 行m 列的矩阵。

求矩阵中上下对称且左右对称的正方形子矩阵的个数。

样例输入

5 5
4 2 4 4 4 
3 1 4 4 3 
3 5 3 3 3 
3 1 5 3 3 
4 2 1 2 4 

样例输出

27

思路

将矩阵上下颠倒一次，左右颠倒一次，再求出每个矩阵的每个区间矩阵的hash值，二分正方形的个数，比较hash值就可以判断是否是上下左右对称。
一个合法正方形满足在该正方形下的正方形都是合法的

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define zhi1 1000000007ull
#define zhi2 1000000009ull
#define ull unsigned long long 

using namespace std;

int n, m, tot, lx, ly, hash1, hash2, hash3, ans, l, r;
ull hasha[1005][1005], hashb[1005][1005], hashc[1005][1005];
ull a[1005][1005], b[1005][1005], c[1005][1005], t1[10005], t2[10005];

bool check(int rx, int ry, int dis) {
    
    
	lx = rx - dis + 1;
	ly = ry - dis + 1;
	hash1 = hasha[rx][ry] - hasha[rx][ly - 1] * t1[dis] - hasha[lx - 1][ry] * t2[dis] + hasha[lx - 1][ly - 1] * t1[dis] * t2[dis];//求出原矩阵的hash值
	
	int tmp = rx;
	rx = n - (rx - dis);
	lx = rx - dis + 1;
	ly = ry - dis + 1;
	hash2 = hashc[rx][ry] - hashc[rx][ly - 1] * t1[dis] - hashc[lx - 1][ry] * t2[dis] + hashc[lx - 1][ly - 1] * t1[dis] * t2[dis];//求出上下折叠后矩阵的hash值
	
	rx = tmp;
	ry = m - (ry - dis);
	lx = rx - dis + 1;
	ly = ry - dis + 1;
	hash3 = hashb[rx][ry] - hashb[rx][ly - 1] * t1[dis] - hashb[lx - 1][ry] * t2[dis] + hashb[lx - 1][ly - 1] * t1[dis] * t2[dis];//求出左右折叠矩阵的hash值
	
	if (hash1 == hash2 && hash1 == hash3) return 1;
	return 0;
}

int main()
{
    
    
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i)
	{
    
    
		for(int j = 1; j <= m; ++j)
		{
    
    
			scanf("%d", &a[i][j]);
			b[i][m - j + 1] = a[i][j];
			c[n - i + 1][j] = a[i][j];
		}
	}
	t1[0] = 1ull;
	for(int i = 1; i <= n; ++i)
		t1[i] = t1[i - 1] * zhi1;
	t2[0] = 1ull;
	for(int i = 1; i <= m; ++i)
		t2[i] = t2[i - 1] * zhi2;
	for(int i = 1; i <= n; ++i) {
    
    
		for(int j = 1; j <= m; ++j)
		{
    
    
			hasha[i][j] = hasha[i][j - 1] * zhi1 + a[i][j];
			hashb[i][j] = hashb[i][j - 1] * zhi1 + b[i][j];
			hashc[i][j] = hashc[i][j - 1] * zhi1 + c[i][j];
		}
		t1[i] = t1[i - 1] * zhi1;
	}
	for(int i = 1; i <= n; ++i) 
		for(int j = 1; j <= m; ++j)
		{
    
    
			hasha[i][j] += hasha[i - 1][j] * zhi2;
			hashb[i][j] += hashb[i - 1][j] * zhi2;
			hashc[i][j] += hashc[i - 1][j] * zhi2;
		}
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j) 
		{
    
    
			ans = 0;
			l = 1;
			r = min(min(i, n - i + 1), min(j, m - j + 1));
			while(l <= r)	
			{
    
    
				int mid = (l + r) >> 1;
				if(i - mid + 1 < 1 || i + mid - 1 > n || j - mid + 1 < 1 || j + mid - 1 > m) {
    
    
					r = mid - 1;
					continue;
				}
				if(check(i + mid - 1, j + mid - 1, mid * 2 - 1)) {
    
    
					ans = mid;
					l = mid + 1;
				} 
				else r = mid - 1;
			}//正方形长度为奇数时
			tot += ans;
			
			ans = 0;
			l = 1;
			r = min(min(i, n - i), min(j, m - j));
			while(l <= r)
			{
    
    
				int mid = (l + r) >> 1;
				if(i - mid + 1 < 1 || i + mid > n || j - mid + 1 < 1 || j + mid > m) {
    
    
					r = mid - 1;
					continue;
				}
				if(check(i + mid, j + mid, mid * 2)) {
    
    
					ans = mid;
					l = mid + 1;
				}
				else r = mid - 1;
			}//正方形长度为偶数时
			tot += ans;
		}
		printf("%d", tot);
}