2=>1,3,4,5,6,7,9
3=>2,4,5,6,8
4=>1,2,3,5,7,8,9
5=>1,2,3,4,6,7,8,9
6=>1,2,3,5,7,8,9
7=>2,4,5,6,8
8=>1,3,4,5,6,7,9
9=>2,4,5,6,8
不错，这种方法好是好，只是有一种情况没考虑吧。
比如，前面如果已经经过5的话，后面4可以直接跳到6的
参考6#可行方案的2516图
路径{8,5,7,4,6}
待我写个程序来
#include <stdio.h>
int mask = {1,0,0,0,0,0,0,0,0,0};
int next =
{0,0,0,0,0,0,0,0,0,0,
5,2,4,5,6,8,0,0,0,0,
7,1,3,4,5,6,7,9,0,0,
5,2,4,5,6,8,0,0,0,0,
7,1,2,3,5,7,8,9,0,0,
8,1,2,3,4,6,7,8,9,0,
7,1,2,3,5,7,8,9,0,0,
5,2,4,5,6,8,0,0,0,0,
7,1,3,4,5,6,7,9,0,0,
5,2,4,5,6,8,0,0,0,0,
};
int counter = {0,0,0,0,0,0,0,0,0,0};
int graph = {0,0,0,0,0,0,0,0,0,0};
int nextNumber = 0;
int graphPostion = 0;
void first( void )
{
int i;
for (i = 1; i <= 9 ; i ++)
{
mask = 1;
graph = i;
circle( graphPostion + 1 );
mask = 0;
}
}
void circle( int graphPostion )
{
int i;
int noPostion = 1;
int n = graph;
for (i = 1; i <= next; i ++)
{
if (mask] == 0)
{
noPostion = 0;
mask] = 1;
graph = next;
counter ++;
circle( graphPostion + 1 );
mask] = 0;
}
}
if (noPostion)
{
printf("find: ");
for (i = 0; i < graphPostion; i ++)
{
printf("%1d", graph);
}
printf("\n");
}
}
int main ( void )
{
int i;
first( );
for (i = 4; i <= 9; i ++)
printf("%d: %d\n", i, counter);
return 0;
}
4: 1400
5: 5328
6: 16032
7: 35328
8: 49536
9: 32256
PointsToBeChecked[{{a_,b_},{c_,d_}}]:=Module[{g=GCD},If,{}]];
n=3;ddd=Flatten,1];
Table[{ii,Product,ddd=Select&/@Complement-1,1],ii],{ii,ddd}],1],Length]]]==0&];Length},{ii,2,n^2}]对算法做了改进，反而跟无心人的答案对上了。
{2, 72, 56},
{3, 504, 304},
{4, 3024, 1400},
{5, 15120, 5328},
{6, 60480, 16032},
{7, 181440, 35328},
{8, 362880, 49536},
{9, 362880, 32256}}
算得4*4的情况：
2,240,172
3,3360,1696
4,43680,15580
5,524160,132264
6,5765760,1029232
7,57657600,7286016
后面就算不动了
:)
证明俺算法没错了哦
取出所有的排列,然后一个一个地判定!
看了下无心人的程序，能计算#6给出的合法情况吗？
5->2->7->3->4
因为5是已经标记过的，所以可以从7直接到3了。