【ybt高效进阶2-2-5】子正方形

子正方形

题目链接：ybt高效进阶2-2-5

题目大意

有两个正方形矩阵，问你这两个矩形的最大公共子正方形矩阵的边长。

思路

这道题矩阵大小才 $50$ 一下，我们可以直接用 hash 暴搜。

先枚举两个矩阵所选的矩阵的右下角，然后再二分枚举矩阵大小。
然后取最大值即可。

代码

#include<cstdio>
#include<iostream>
#define mi1 97ull
#define mi2 103ull
#define ull unsigned long long

using namespace std;

int n, a[51][51], b[51][51], l, r, mid, ans, answer;
ull a_hash[51][51], b_hash[51][51], t1[51], t2[51];

bool compair(int sx, int sy, int tx, int ty, int mid) {
    
    //看两个矩阵是不是
	return (a_hash[sx][sy] - a_hash[sx][sy - mid] * t1[mid] - a_hash[sx - mid][sy] * t2[mid] + a_hash[sx - mid][sy - mid] * t1[mid] * t2[mid])
			==
			(b_hash[tx][ty] - b_hash[tx][ty - mid] * t1[mid] - b_hash[tx - mid][ty] * t2[mid] + b_hash[tx - mid][ty - mid] * t1[mid] * t2[mid])
			;
}

int main() {
    
    
	scanf("%d", &n);
	
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			scanf("%d", &a[i][j]);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			scanf("%d", &b[i][j]);
	
	t1[0] = t2[0] = 1ull;
	for (int i = 1; i <= n; i++) t1[i] = t1[i - 1] * mi1;
	for (int i = 1; i <= n; i++) t2[i] = t2[i - 1] * mi2;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) {
    
    
			a_hash[i][j] = a_hash[i][j - 1] * mi1 + a[i][j];
			b_hash[i][j] = b_hash[i][j - 1] * mi1 + b[i][j];
		}
	for (int i = 1; i <= n; i++)//得出两个矩阵的hash值
		for (int j = 1; j <= n; j++) {
    
    
			a_hash[i][j] += a_hash[i - 1][j] * mi2;
			b_hash[i][j] += b_hash[i - 1][j] * mi2;
		}
	
	for (int sx = 1; sx <= n; sx++)//枚举第一个矩阵右下角选哪里
		for (int sy = 1; sy <= n; sy++)
			for (int tx = 1; tx <= n; tx++)//枚举第二个矩阵右下角选哪里
				for (int ty = 1; ty <= n; ty++) {
    
    
					l = 0;
					r = min(min(sx, sy), min(tx, ty));
					ans = 0;
					while (l <= r) {
    
    //二分出最大的大小
						mid = (l + r) >> 1;
						if (compair(sx, sy, tx, ty, mid)) {
    
    
							ans = mid;
							l = mid + 1;
						}
						else r = mid - 1;
					}
					answer = max(answer, ans);
				} 
	
	printf("%d", answer);
	
	return 0;
}