思路：将图1和图2分别平移到另一张图上，通过遍历求得相同数量，取最大值。

#include <bits/stdc++.h>
using namespace std;
const int N=51;
int a[N][N],b[N][N];
int n1,m1,n2,m2;
int main()
{
	//freopen("B3788.in","r",stdin);
	//freopen("B3788.out","w",stdout);
	memset (a,0x3f,sizeof a);
	memset (b,0x3f,sizeof b);
	scanf("%d%d",&n1,&m1);
	for (int i=1;i<=n1;i++)
	{
		for (int j=1;j<=m1;j++)
		{
			scanf("%d",&a[i][j]);
		}
	}
	scanf("%d%d",&n2,&m2);
	for (int i=1;i<=n2;i++)
	{
		for (int j=1;j<=m2;j++)
		{
			scanf("%d",&b[i][j]);
		}
	}
	int ans=0;
	int x1=1,y1=1;
	while (x1*y1<=n1*m1)
	{
		int cnt=0;
		for (int i=x1;i<=n1;i++)
		{
			for (int j=y1;j<=m1;j++)
			{
				if (a[i][j]==b[i-x1+1][j-y1+1]) cnt++;
			}
		}
		ans=max(ans,cnt);
		if (y1==m1&&x1!=n1) x1++,y1=1;
		else y1++;
	}
	int x2=1,y2=1;
	while (x2*y2<=n2*m2)
	{
		int cnt=0;
		for (int i=x2;i<=n2;i++)
		{
			for (int j=y2;j<=m2;j++)
			{
				if (b[i][j]==a[i-x2+1][j-y2+1]) cnt++;
			}
		}
		ans=max(ans,cnt);
		if (y2==m2&&x2!=n2) x2++,y2=1;
		else y2++;
	}
	printf("%d",ans);
	return 0;
}