2023-11-18：用go语言，如果一个正方形矩阵上下对称并且左右对称，对称的意思是互为镜像，
那么称这个正方形矩阵叫做神奇矩阵。
比如 :
1 5 5 1
6 3 3 6
6 3 3 6
1 5 5 1
这个正方形矩阵就是神奇矩阵。
给定一个大矩阵n*m，返回其中神奇矩阵的数目。
1 <= n,m <= 1000。
来自左程云。
答案2023-11-18：
go，c ，c的代码用灵捷3.5编写，go和c 有修改。
具体步骤如下：
1.通过输入获取大矩阵的大小n和m。
2.将输入的数据按行列填充到数组arr中。
3.根据行遍历，对每一行调用manacher函数进行回文串的预处理。该函数会在rp数组中保存每个位置向右的回文长度。
4.根据列遍历，对每一列调用manacher函数进行回文串的预处理。该函数会在cp数组中保存每个位置向下的回文长度。
5.遍历所有内部的行和列，计算每个位置上、下、左、右四个方向上的回文长度，并取其最小值作为当前位置的enlarge值。
6.统计enlarge数组中每个奇数行、奇数列位置的值除以2的结果，作为神奇矩阵的数量。
7.统计enlarge数组中每个偶数行、偶数列位置的值减去1后除以2的结果，再累加到神奇矩阵的数量。
8.返回神奇矩阵的数量作为结果。
总的时间复杂度：O(n * m * log(min(n, m)))，其中n为矩阵的行数，m为矩阵的列数。主要耗时的是manacher函数的预处理过程，而manacher函数的时间复杂度为O(log(min(n, m)))。
总的额外空间复杂度：O(n * m)，需要额外的数组保存回文长度。
go完整代码如下：
代码语言：javascript复制package main
import (
"fmt"
)
const MAXN = 1001
var log2 [(MAXN<<1 | 1) 1]int
var arr [MAXN<<1 | 1][MAXN<<1 | 1]int
var rp [MAXN<<1 | 1][MAXN<<1 | 1]int
var cp [MAXN<<1 | 1][MAXN<<1 | 1]int
var enlarge [MAXN<<1 | 1][MAXN<<1 | 1]int
var rmq [MAXN<<1 | 1][MAXN<<1 | 1]int
var s [MAXN<<1 | 1]int
var p [MAXN<<1 | 1]int
var n, m int
func init() {
for k, j := 0, 1; j <= (MAXN<<1 | 1); j {
if 1<<(k 1) <= j {
k
}
log2[j] = k
}
}
func main() {
inputs := []int{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}
ii := 0
n = inputs[ii]
ii
m = inputs[ii]
ii
for i, r := 0, 1; i < n; i, r = i 1, r 2 {
for j, c := 0, 1; j < m; j, c = j 1, c 2 {
arr[r][c] = inputs[ii]
ii
}
}
n = n*2 1
m = m*2 1
fmt.Println(number())
}
func number() int {
for row := 0; row < n; row {
manacher(row, 0, 0, 1)
}
for col := 0; col < m; col {
manacher(0, col, 1, 0)
}
for row := 1; row < n-1; row {
rowRmq(row)
for col := 1; col < m-1; col {
l := 1
r := min(min(row 1, n-row), min(col 1, m-col))
find := 1
for l <= r {
m := (l r) / 2
if query(col-m 1, col m-1) >= m {
find = m
l = m 1
} else {
r = m - 1
}
}
enlarge[row][col] = find
}
}
for col := 1; col < m-1; col {
colRmq(col)
for row := 1; row < n-1; row {
l := 1
r := min(min(row 1, n-row), min(col 1, m-col))
find := 1
for l <= r {
m := (l r) / 2
if query(row-m 1, row m-1) >= m {
find = m
l = m 1
} else {
r = m - 1
}
}
enlarge[row][col] = min(enlarge[row][col], find)
}
}
ans := 0
for row := 1; row < n-1; row = 2 {
for col := 1; col < m-1; col = 2 {
ans = enlarge[row][col] / 2
}
}
for row := 2; row < n-1; row = 2 {
for col := 2; col < m-1; col = 2 {
ans = (enlarge[row][col] - 1) / 2
}
}
return ans
}
func manacher(row int, col int, radd int, cadd int) {
limit := 0
for r, c := row, col; r < n && c < m; r, c = r radd, c cadd {
s[limit] = arr[r][c]
limit
}
C := -1
R := -1
for i := 0; i < limit; i {
p[i] = R
if i < R {
p[i] = min(p[2*C-i], R-i)
} else {
p[i] = 1
}
for i p[i] < limit && i-p[i] > -1 && s[i p[i]] == s[i-p[i]] {
p[i]
}
if i p[i] > R {
R = i p[i]
C = i
}
}
var fill *[2003][2003]int
if cadd == 1 {
fill = &rp
} else {
fill = &cp
}
for i, r, c := 0, row, col; i < limit; i {
fill[r][c] = p[i]
r = radd
c = cadd
}
}
func rowRmq(row int) {
for i := 0; i < m; i {
rmq[i][0] = cp[row][i]
}
for j := 1; (1 << j) <= m; j {
for i := 0; i (1<<j)-1 < m; i {
rmq[i][j] = min(rmq[i][j-1], rmq[i (1<<(j-1))][j-1])
}
}
}
func colRmq(col int) {
for i := 0; i < n; i {
rmq[i][0] = rp[i][col]
}
for j := 1; (1 << j) <= n; j {
for i := 0; i (1<<j)-1 < n; i {
rmq[i][j] = min(rmq[i][j-1], rmq[i (1<<(j-1))][j-1])
}
}
}
func query(l int, r int) int {
k := log2[r-l 1]
return min(rmq[l][k], rmq[r-(1<<k) 1][k])
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
在这里插入图片描述
c 完整代码如下：
代码语言：javascript复制#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1001;
int log22[(MAXN << 1 | 1) 1];
int arr[MAXN << 1 | 1][MAXN << 1 | 1];
int rp[MAXN << 1 | 1][MAXN << 1 | 1];
int cp[MAXN << 1 | 1][MAXN << 1 | 1];
int enlarge[MAXN << 1 | 1][MAXN << 1 | 1];
int rmq[MAXN << 1 | 1][MAXN << 1 | 1];
int s[MAXN << 1 | 1];
int p[MAXN << 1 | 1];
int n, m;
void manacher(int row, int col, int radd, int cadd);
int number();
void rowRmq(int row);
void colRmq(int col);
int query(int l, int r);
int min(int a, int b);
void init() {
for (int k = 0, j = 1; j <= (MAXN << 1 | 1); j ) {
if (1 << (k 1) <= j) {
k ;
}
log22[j] = k;
}
}
int number() {
for (int row = 0; row < n; row ) {
manacher(row, 0, 0, 1);
}
for (int col = 0; col < m; col ) {
manacher(0, col, 1, 0);
}
for (int row = 1; row < n - 1; row ) {
rowRmq(row);
for (int col = 1; col < m - 1; col ) {
int l = 1;
int r = min(min(row 1, n - row), min(col 1, m - col));
int find = 1;
while (l <= r) {
int mid = (l r) / 2;
if (query(col - mid 1, col mid - 1) >= mid) {
find = mid;
l = mid 1;
}
else {
r = mid - 1;
}
}
enlarge[row][col] = find;
}
}
for (int col = 1; col < m - 1; col ) {
colRmq(col);
for (int row = 1; row < n - 1; row ) {
int l = 1;
int r = min(min(row 1, n - row), min(col 1, m - col));
int find = 1;
while (l <= r) {
int mid = (l r) / 2;
if (query(row - mid 1, row mid - 1) >= mid) {
find = mid;
l = mid 1;
}
else {
r = mid - 1;
}
}
enlarge[row][col] = min(enlarge[row][col], find);
}
}
int ans = 0;
for (int row = 1; row < n - 1; row = 2) {
for (int col = 1; col < m - 1; col = 2) {
ans = enlarge[row][col] / 2;
}
}
for (int row = 2; row < n - 1; row = 2) {
for (int col = 2; col < m - 1; col = 2) {
ans = (enlarge[row][col] - 1) / 2;
}
}
return ans;
}
void manacher(int row, int col, int radd, int cadd) {
int limit = 0;
for (int r = row, c = col; r < n && c < m; r = radd, c = cadd) {
s[limit] = arr[r][c];
limit ;
}
int C = -1;
int R = -1;
for (int i = 0; i < limit; i ) {
p[i] = R;
if (i < R) {
p[i] = min(p[2 * C - i], R - i);
}
else {
p[i] = 1;
}
while (i p[i] < limit && i - p[i] > -1 && s[i p[i]] == s[i - p[i]]) {
p[i] ;
}
if (i p[i] > R) {
R = i p[i];
C = i;
}
}
int(*fill)[2003];
if (cadd == 1) {
fill = rp;
}
else {
fill = cp;
}
for (int i = 0, r = row, c = col; i < limit; i ) {
fill[r][c] = p[i];
r = radd;
c = cadd;
}
}
void rowRmq(int row) {
for (int i = 0; i < m; i ) {
rmq[i][0] = cp[row][i];
}
for (int j = 1; (1 << j) <= m; j ) {
for (int i = 0; i (1 << j) - 1 < m; i ) {
rmq[i][j] = min(rmq[i][j - 1], rmq[i (1 << (j - 1))][j - 1]);
}
}
}
void colRmq(int col) {
for (int i = 0; i < n; i ) {
rmq[i][0] = rp[i][col];
}
for (int j = 1; (1 << j) <= n; j ) {
for (int i = 0; i (1 << j) - 1 < n; i ) {
rmq[i][j] = min(rmq[i][j - 1], rmq[i (1 << (j - 1))][j - 1]);
}
}
}
int query(int l, int r) {
int k = log22[r - l 1];
return min(rmq[l][k], rmq[r - (1 << k) 1][k]);
}
int min(int a, int b) {
if (a < b) {
return a;
}
return b;
}
int main() {
init();
int inputs[] = { 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 };
int ii = 0;
n = inputs[ii ];
m = inputs[ii ];
for (int i = 0, r = 1; i < n; i , r = 2) {
for (int j = 0, c = 1; j < m; j , c = 2) {
arr[r][c] = inputs[ii ];
}
}
n = n * 2 1;
m = m * 2 1;
cout << number() << endl;
return 0;
}
在这里插入图片描述
c完整代码如下：
代码语言：javascript复制#include <stdio.h>
#define MAXN 1001
int log2Arr[(MAXN << 1 | 1) 1];
int arr[MAXN << 1 | 1][MAXN << 1 | 1];
int rp[MAXN << 1 | 1][MAXN << 1 | 1];
int cp[MAXN << 1 | 1][MAXN << 1 | 1];
int enlarge[MAXN << 1 | 1][MAXN << 1 | 1];
int rmq[MAXN << 1 | 1][MAXN << 1 | 1];
int s[MAXN << 1 | 1];
int p[MAXN << 1 | 1];
int n, m;
void init() {
int k = 0;
for (int j = 1; j <= (MAXN << 1 | 1); j ) {
if (1 << (k 1) <= j) {
k ;
}
log2Arr[j] = k;
}
}
int min(int a, int b) {
return (a < b) ? a : b;
}
void manacher(int row, int col, int radd, int cadd) {
int limit = 0;
for (int r = row, c = col; r < n && c < m; r = radd, c = cadd) {
s[limit] = arr[r][c];
limit ;
}
int C = -1;
int R = -1;
for (int i = 0; i < limit; i ) {
p[i] = R;
if (i < R) {
p[i] = min(p[2 * C - i], R - i);
}
else {
p[i] = 1;
}
while (i p[i] < limit && i - p[i] > -1 && s[i p[i]] == s[i - p[i]]) {
p[i] ;
}
if (i p[i] > R) {
R = i p[i];
C = i;
}
}
int(*fill)[2003];
if (cadd == 1) {
fill = rp;
}
else {
fill = cp;
}
for (int i = 0, r = row, c = col; i < limit; i ) {
fill[r][c] = p[i];
r = radd;
c = cadd;
}
}
void rowRmq(int row) {
for (int i = 0; i < m; i ) {
rmq[i][0] = cp[row][i];
}
for (int j = 1; (1 << j) <= m; j ) {
for (int i = 0; i (1 << j) - 1 < m; i ) {
rmq[i][j] = min(rmq[i][j - 1], rmq[i (1 << (j - 1))][j - 1]);
}
}
}
void colRmq(int col) {
for (int i = 0; i < n; i ) {
rmq[i][0] = rp[i][col];
}
for (int j = 1; (1 << j) <= n; j ) {
for (int i = 0; i (1 << j) - 1 < n; i ) {
rmq[i][j] = min(rmq[i][j - 1], rmq[i (1 << (j - 1))][j - 1]);
}
}
}
int query(int l, int r) {
int k = log2Arr[r - l 1];
return min(rmq[l][k], rmq[r - (1 << k) 1][k]);
}
int number() {
for (int row = 0; row < n; row ) {
manacher(row, 0, 0, 1);
}
for (int col = 0; col < m; col ) {
manacher(0, col, 1, 0);
}
for (int row = 1; row < n - 1; row ) {
rowRmq(row);
for (int col = 1; col < m - 1; col ) {
int l = 1;
int r = min(min(row 1, n - row), min(col 1, m - col));
int find = 1;
while (l <= r) {
int mid = (l r) / 2;
if (query(col - mid 1, col mid - 1) >= mid) {
find = mid;
l = mid 1;
}
else {
r = mid - 1;
}
}
enlarge[row][col] = find;
}
}
for (int col = 1; col < m - 1; col ) {
colRmq(col);
for (int row = 1; row < n - 1; row ) {
int l = 1;
int r = min(min(row 1, n - row), min(col 1, m - col));
int find = 1;
while (l <= r) {
int mid = (l r) / 2;
if (query(row - mid 1, row mid - 1) >= mid) {
find = mid;
l = mid 1;
}
else {
r = mid - 1;
}
}
enlarge[row][col] = min(enlarge[row][col], find);
}
}
int ans = 0;
for (int row = 1; row < n - 1; row = 2) {
for (int col = 1; col < m - 1; col = 2) {
ans = enlarge[row][col] / 2;
}
}
for (int row = 2; row < n - 1; row = 2) {
for (int col = 2; col < m - 1; col = 2) {
ans = (enlarge[row][col] - 1) / 2;
}
}
return ans;
}
int main() {
init();
int inputs[] = { 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 };
int ii = 0;
n = inputs[ii];
ii ;
m = inputs[ii];
ii ;
for (int i = 0, r = 1; i < n; i , r = 2) {
for (int j = 0, c = 1; j < m; j , c = 2) {
arr[r][c] = inputs[ii];
ii ;
}
}
n = n * 2 1;
m = m * 2 1;
printf("%dn", number());
return 0;
}
在这里插入图片描述
