标签:Hash 正方形 int ybtoj ll h1 h2 55 xr
子正方形
Description–
高效进阶「字符串算法」第2章 Hash 和 Hash 表课堂过关 例题5
代码–
#include <iostream>
#include <cstdio>
#define ull long long
using namespace std;
int n, ans, a[55][55], b[55][55];
ull t1[55], t2[55], h1[55][55], h2[55][55];
bool xyn(int x, int y, int xr, int yr, int ll) //求这两个矩阵的哈希值
{
ull aa, bb;
aa = h1[x][y] - h1[x - ll][y] * t2[ll] - h1[x][y - ll] * t1[ll] + h1[x - ll][y - ll] * t1[ll] * t2[ll];
bb = h2[xr][yr] - h2[xr - ll][yr] * t2[ll] - h2[xr][yr - ll] * t1[ll] + h2[xr - ll][yr - ll] * t1[ll] * t2[ll];
return aa == bb;
}
void hll(int x, int y, int xr, int yr) //二分
{
int l = 0, r = min(min(x, y), min(xr, yr));
while (l <= r)
{
int mid = (l + r + 1) >> 1;
if (xyn(x, y, xr, yr, mid))
l = mid + 1, ans = max(ans, mid);
else r = mid - 1;
}
}
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] = 1;
for (int i = 1; i <= n; ++i)
t1[i] = t1[i - 1] * 131, t2[i] = t2[i - 1] * 61;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
{
h1[i][j] = h1[i][j - 1] * 131 + a[i][j];
h2[i][j] = h2[i][j - 1] * 131 + b[i][j];
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
{
h1[i][j] += h1[i - 1][j] * 61;
h2[i][j] += h2[i - 1][j] * 61;
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
for (int k = 1; k <= n; ++k)
for (int l = 1; l <= n; ++l)
if (a[i][j] == b[k][l])
hll(i, j, k, l);
printf("%d", ans);
return 0;
}
标签:Hash,正方形,int,ybtoj,ll,h1,h2,55,xr 来源: https://blog.csdn.net/qq_43654542/article/details/116533956
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