//皇后问题最快的解法（C语言）八皇后、十六皇后时间在毫秒级
#include <windows.h>
# include<stdio.h>
# include<stdlib.h>
# include<time.h>
#define O 134775813
// TODO (xubo18056052430#1#): PB10210016 徐波
LARGE_INTEGER lv,lv1,lv2,lv3,lv4;
double totaltime,secondsPerTick;
int step;
/***********************************************************************************************/
int dn(int n) //分组，根据经验值产生，网搜得到--取下列值是比较好，返回不同的值对实验结果影响很大
{
if (n < 100)
return n;
else if (n < 1000)
return 40;
else if (n < 10000)
return 60;
else if (n < 100000)
return 90;
else
return 150;
}
/***********************************************************************************************/
/***********************************************************************************************/
int rand1 = (unsigned)time(NULL); //rand1产生随机数
int rand2(int range) //rand1线性同余随机数产生器
{
unsigned int x = 0x7fffffff, m, c = 1, a =O;
m = x + 1;
rand1 = (rand1 * a + 1)%m;
double r =(double) rand1 / x;
int r2 = (int)(r * range);
return r2;
}
/***********************************************************************************************/
/***********************************************************************************************/
void swap(int &a, int &b){ //交换两值
int c;
c = a;a = b;b = c;
}
/***********************************************************************************************/
/***********************************************************************************************/
int chongtu(int n, int line1[], int line2[]){ //冲突
int xubo = 0;
for(int i = 0; i < n + n; i ++){
if(line1[i] > 1){xubo = xubo + (line1[i] * (line1[i] - 1)) / 2;}
if(line2[i] > 1){xubo = xubo + (line2[i] * (line2[i] - 1)) / 2;}
}
return xubo;
}
/***********************************************************************************************/
/***********************************************************************************************/
int improve(int i, int j, int n, int q[], int line1[], int line2[]) //判断q[i]和q[j])交换后冲突的个数是否会减少
{
int xubo = 0;
xubo = xubo + (1 - line1[i + q[i]]);
xubo = xubo + (1 - line2[i - q[i] + n - 1]);
xubo = xubo + (1 - line1[j + q[j]]);
xubo = xubo + (1 - line2[j - q[j] + n - 1]);
xubo = xubo + line1[i + q[j]];
xubo = xubo + line2[i - q[j] + n - 1];
xubo = xubo + line1[j + q[i]];
xubo = xubo + line2[j - q[i] + n - 1];
if(i + q[i] == j + q[j] || i - q[i] == j - q[j])
xubo = xubo + 2;
return xubo;
}
/***********************************************************************************************/
/***********************************************************************************************/
void csh(int n, int m, int q[], int line1[], int line2[])
{
int i, end, rand3;
bool *l1 = (bool *)malloc((n + n) * sizeof(bool));
bool *l2 = (bool *)malloc((n + n) * sizeof(bool));
for(i = 0; i < n; i ++)
{
q[i] = i;
line1[i] = 0;
line2[i] = 0;
l1[i] = false;
l2[i] = false;
}
for(i = n; i < n + n; i ++)
{
line1[i] = 0;
line2[i] = 0;
l1[i] = false;
l2[i] = false;
}
for(i = 0, end = n; i < m; i ++, end --) //初始化（0 ~ m - 1）；前m行不会此时不会产生冲突
{ //初始化后横竖无冲突，只要保证两斜线无冲突即可
do
{
rand3 = i + rand2(end);
}while(l1[i + q[rand3]] || l2[i - q[rand3] + n - 1]);
swap(q[i], q[rand3]);
line1[i + q[i]] ++;
line2[i - q[i] + n - 1] ++;
l1[i + q[i]] = true;
l2[i - q[i] + n - 1] = true;
}
for(i = m, end = n - m; i < n; i ++, end--) //初始化line
{
rand3 = i + rand2(end);
swap(q[i], q[rand3]);
line1[i + q[i]] ++;
line2[i - q[i] + n - 1] ++;
}
free(l1);
free(l2);
}
/***********************************************************************************************/
/***********************************************************************************************/
void work(int n, int m, int q[])
{
QueryPerformanceFrequency( &lv );
secondsPerTick = 1000000.0 / lv.QuadPart;
int temp, reduce, i, j;
step = 0;
int *line1 = (int *)malloc((n + n) * sizeof(int));
int *line2 = (int *)malloc((n + n) * sizeof(int));
QueryPerformanceCounter( &lv3 );
csh(n, m, q, line1, line2);
QueryPerformanceCounter( &lv4 );
totaltime = secondsPerTick * (lv4.QuadPart-lv3.QuadPart);
printf("\n初始化时间为：%e s ", totaltime/1000000); //初始化时间
temp = chongtu(n, line1, line2);
//printf("冲突次数= %d\n", temp);
while(temp > 0){
reduce = 0;
for(i = m; i < n; i ++){
if(line1[i + q[i]] == 1 && line2[i - q[i] + n - 1] == 1)
continue;
else{
for(j = 0; j < n; j ++){
if(i != j){
reduce = improve(i, j, n, q, line1, line2);
if(reduce < 0)
break;
}
}
if(reduce < 0) //可交换q[i]和q[j]，使总的冲突减少
break;
}
}
if(reduce < 0){
step ++;
line1[i + q[i]] --;
line2[i - q[i] + n - 1] --;
line1[j + q[j]] --;
line2[j - q[j] + n - 1] --;
swap(q[i], q[j]);
line1[i + q[i]] ++;
line2[i - q[i] + n - 1] ++;
line1[j + q[j]] ++;
line2[j - q[j] + n - 1] ++;
temp = temp + reduce;
}
else
{
csh(n, m, q, line1, line2);
temp = chongtu(n, line1, line2);
}
}
}
/***********************************************************************************************/
/***********************************************************************************************/
size_t **NQueen(size_t q_xubober) //实验要求size_t **NQueen(size_t Queen_xubober)
{
LARGE_INTEGER lv,lv1,lv2;
double totaltime,secondsPerTick;
QueryPerformanceFrequency( &lv );
secondsPerTick = 1000000.0 / lv.QuadPart;
unsigned int **ee = (unsigned int **)malloc(4 * sizeof(unsigned int *)), loop = 3, i, j;
for(i = 1; i < 4; i ++)
ee[i] = (unsigned int *)malloc((q_xubober + 1) * sizeof(unsigned int));
int *q = (int *)malloc(q_xubober * sizeof(int));
int m = q_xubober - dn(q_xubober);
printf("\n问题大小N= %d m= %d\n*************************************************************************\n", q_xubober, m);
for(i = 1; i < 4; i ++){
QueryPerformanceCounter( &lv1 );
work(q_xubober, m, q);
QueryPerformanceCounter( &lv2 );
totaltime = secondsPerTick * (lv2.QuadPart-lv1.QuadPart);
printf("第%d次成功运行时间= %e s \n", i, totaltime/1000000);
printf("\n*************************************************************************\n");
for(j = 0; j < q_xubober; j ++)
ee[i][j + 1] = q[j] + 1;
}
return ee;
}
/***********************************************************************************************/
/***********************************************************************************************/
int main(){
unsigned int i, j, k = 0, **ee = NULL;
int n;
printf("请输入皇后问题N的大小:\n");
scanf("%d", &n);
printf("\n————————————————运行结果————————————————\n");
if(n<=3){
printf("\错误！没有解！");
exit(1); //输入小于4的数报错跳出
}
if(n>=1000000000003){
printf("\错误！没有解！");
exit(1); //输入大于的数报错跳出
}
ee = NQueen(n);
if(n<=100){ //如果输入的n小于等于100，者输出皇后摆放位置序列，否者不输出，因为n大序列长
for(i = 1; i <=3; i++) //输出三个满足问题要求的皇后摆放位置序列try[1..3][1…n]。
{
printf("\n\n第%d组解皇后摆放的位置序列位：\n",i);
for(int j = 1; j < n + 1; j ++){
for(k = 1; k < n + 1; k ++){
if(k ==ee[i][j]){
//printf("%d ",k); //只输出列
printf("（%d，%d） ",j,k);} //输出行和列，第i个皇后摆放在(i,try[i])
}
}
}
}
printf("\n\n————————————————运行结果————————————————\n\n\n");
system("pause");
return 0;
}
/***********************************************************************************************/
// TODO (xubo18056052430#1#): PB10210016 徐波
