求16宫格C程序算法
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解决时间 2021-02-24 19:24
- 提问者网友：我的未来我做主
- 2021-02-24 01:33
如题，想了很久，用了最笨的办法，但需要时间太长（n天），希望有高手提供精辟的算法，在短时间内求解所有16宫格，要求用C语言或C++语言，所给代码可以运行出结果。ps:16宫格与9宫格类似，在4*4矩阵中，4行与4列与2对角线这10个各自的累加值均相等，且这16个数是从1~16选出，不可重复，即每个数用且仅用一次。我的积分本来就不多，只能提供一点咯，望高手指点~不胜感激~
最佳答案
- 五星知识达人网友：一秋
- 2021-02-24 02:28
N阶幻方构造#include
using namespace std;//幻方构造(l!=2)
#define MAXN 100void dllb(int l,int si,int sj,int sn,int d[][MAXN]){
int n,i=0,j=l/2;
for (n=1;n<=l*l;n++){
d[i+si][j+sj]=n+sn;
if (n%l){
i=(i)?(i-1):(l-1);
j=(j==l-1)?0:(j+1);
}
else
i=(i==l-1)?0:(i+1);
}
}void magic_odd(int l,int d[][MAXN]){
dllb(l,0,0,0,d);
}void magic_4k(int l,int d[][MAXN]){
int i,j;
for (i=0;i
}void magic_other(int l,int d[][MAXN]){
int i,j,t;
dllb(l/2,0,0,0,d);
dllb(l/2,l/2,l/2,l*l/4,d);
dllb(l/2,0,l/2,l*l/2,d);
dllb(l/2,l/2,0,l*l/4*3,d);
for (i=0;i
t=d[i][j],d[i][j]=d[i+l/2][j],d[i+l/2][j]=t;
t=d[l/4][l/4],d[l/4][l/4]=d[l/4+l/2][l/4],d[l/4+l/2][l/4]=t;
for (i=0;i
using namespace std;//幻方构造(l!=2)
#define MAXN 100void dllb(int l,int si,int sj,int sn,int d[][MAXN]){
int n,i=0,j=l/2;
for (n=1;n<=l*l;n++){
d[i+si][j+sj]=n+sn;
if (n%l){
i=(i)?(i-1):(l-1);
j=(j==l-1)?0:(j+1);
}
else
i=(i==l-1)?0:(i+1);
}
}void magic_odd(int l,int d[][MAXN]){
dllb(l,0,0,0,d);
}void magic_4k(int l,int d[][MAXN]){
int i,j;
for (i=0;i
}void magic_other(int l,int d[][MAXN]){
int i,j,t;
dllb(l/2,0,0,0,d);
dllb(l/2,l/2,l/2,l*l/4,d);
dllb(l/2,0,l/2,l*l/2,d);
dllb(l/2,l/2,0,l*l/4*3,d);
for (i=0;i
t=d[i][j],d[i][j]=d[i+l/2][j],d[i+l/2][j]=t;
t=d[l/4][l/4],d[l/4][l/4]=d[l/4+l/2][l/4],d[l/4+l/2][l/4]=t;
for (i=0;i
}void generate(int l,int d[][MAXN]){
if (l%2)
magic_odd(l,d);
else if (l%4==0)
magic_4k(l,d);
else
magic_other(l,d);
}int main() {
int i, j, mat[MAXN][MAXN];
generate(4, mat);
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
printf("%d ", mat[i][j]);
}
putchar('\n');
}
system("pause");
}
using namespace std;//幻方构造(l!=2)
#define MAXN 100void dllb(int l,int si,int sj,int sn,int d[][MAXN]){
int n,i=0,j=l/2;
for (n=1;n<=l*l;n++){
d[i+si][j+sj]=n+sn;
if (n%l){
i=(i)?(i-1):(l-1);
j=(j==l-1)?0:(j+1);
}
else
i=(i==l-1)?0:(i+1);
}
}void magic_odd(int l,int d[][MAXN]){
dllb(l,0,0,0,d);
}void magic_4k(int l,int d[][MAXN]){
int i,j;
for (i=0;i
}void magic_other(int l,int d[][MAXN]){
int i,j,t;
dllb(l/2,0,0,0,d);
dllb(l/2,l/2,l/2,l*l/4,d);
dllb(l/2,0,l/2,l*l/2,d);
dllb(l/2,l/2,0,l*l/4*3,d);
for (i=0;i
t=d[i][j],d[i][j]=d[i+l/2][j],d[i+l/2][j]=t;
t=d[l/4][l/4],d[l/4][l/4]=d[l/4+l/2][l/4],d[l/4+l/2][l/4]=t;
for (i=0;i
using namespace std;//幻方构造(l!=2)
#define MAXN 100void dllb(int l,int si,int sj,int sn,int d[][MAXN]){
int n,i=0,j=l/2;
for (n=1;n<=l*l;n++){
d[i+si][j+sj]=n+sn;
if (n%l){
i=(i)?(i-1):(l-1);
j=(j==l-1)?0:(j+1);
}
else
i=(i==l-1)?0:(i+1);
}
}void magic_odd(int l,int d[][MAXN]){
dllb(l,0,0,0,d);
}void magic_4k(int l,int d[][MAXN]){
int i,j;
for (i=0;i
}void magic_other(int l,int d[][MAXN]){
int i,j,t;
dllb(l/2,0,0,0,d);
dllb(l/2,l/2,l/2,l*l/4,d);
dllb(l/2,0,l/2,l*l/2,d);
dllb(l/2,l/2,0,l*l/4*3,d);
for (i=0;i
t=d[i][j],d[i][j]=d[i+l/2][j],d[i+l/2][j]=t;
t=d[l/4][l/4],d[l/4][l/4]=d[l/4+l/2][l/4],d[l/4+l/2][l/4]=t;
for (i=0;i
}void generate(int l,int d[][MAXN]){
if (l%2)
magic_odd(l,d);
else if (l%4==0)
magic_4k(l,d);
else
magic_other(l,d);
}int main() {
int i, j, mat[MAXN][MAXN];
generate(4, mat);
for (i = 0; i < 4; i++) {
for (j = 0; j < 4; j++) {
printf("%d ", mat[i][j]);
}
putchar('\n');
}
system("pause");
}
全部回答
- 1楼网友：拾荒鲤
- 2021-02-24 02:41
#include
#include
static int pu[9][9]=
{
{0,0,0,7,2,8,0,0,0},
{0,9,0,0,5,1,6,0,0},
{0,0,0,0,6,0,0,8,2},
{3,0,0,8,0,2,7,0,4},
{1,7,4,0,3,0,0,2,0},
{2,8,0,5,0,0,0,3,0},
{0,1,0,3,0,0,2,0,0},
{0,0,7,0,4,6,0,0,5},
{0,0,6,1,0,0,0,4,9} };
int isvalid(const int i, const int j)//验证函数当期i,j坐标是否符合游戏规则，不重复
{
const int n = pu[i][j];
const static int query[] = {0, 0, 0, 3, 3, 3, 6, 6, 6};
int t, u;
for (t = 0; t < 9; t++)
if (t != i && pu[t][j] == n || t != j && pu[i][t] == n)//0-9的数字，每行每列都不能重复
return 0;
for (t = query[i]; t < query[i] + 3; t++) //9个宫的3×3里也不能重复
for (u = query[j]; u < query[j] + 3; u++)
if ((t != i || u != j) && pu[t][u] == n)
return 0;
return 1;
}
void output(void)//输入函数
{
static int n;
cout << "Solution " << ...#include
#include
static int pu[9][9]=
{
{0,0,0,7,2,8,0,0,0},
{0,9,0,0,5,1,6,0,0},
{0,0,0,0,6,0,0,8,2},
{3,0,0,8,0,2,7,0,4},
{1,7,4,0,3,0,0,2,0},
{2,8,0,5,0,0,0,3,0},
{0,1,0,3,0,0,2,0,0},
{0,0,7,0,4,6,0,0,5},
{0,0,6,1,0,0,0,4,9} };
int isvalid(const int i, const int j)//验证函数当期i,j坐标是否符合游戏规则，不重复
{
const int n = pu[i][j];
const static int query[] = {0, 0, 0, 3, 3, 3, 6, 6, 6};
int t, u;
for (t = 0; t < 9; t++)
if (t != i && pu[t][j] == n || t != j && pu[i][t] == n)//0-9的数字，每行每列都不能重复
return 0;
for (t = query[i]; t < query[i] + 3; t++) //9个宫的3×3里也不能重复
for (u = query[j]; u < query[j] + 3; u++)
if ((t != i || u != j) && pu[t][u] == n)
return 0;
return 1;
}
void output(void)//输入函数
{
static int n;
cout << "Solution " << ++n << ":" <
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