數據結構之八皇后問題

用棧實現的八皇后問題

代碼如下：

#include "stdio.h"
#include "stdlib.h"
#include "math.h"

#define _size 100 //棧的大小
#define len 8
/*
	* 八皇后問題
	* 對於兩點(x1,y1),(x2,y2)
	* 不能同行 x1 != x2
	* 不能同列 y1 != y2
	* 不能在對角線上 斜率的絕對值 |k| != 1 |x1 - x2| != |y1 - y2| 
*/

typedef struct{
	int x;
	int y;
}*point,_point;

typedef struct{
	point *base;
	point *top;
	int size;
}*stack,_stack;


int place(_stack st,int x,int y);
void InitStack(stack *S);
void push(stack st,point p);
point pop(stack st);
int length(stack st);
int isEmpty(stack st);
point getPoint(int i,int j);

void InitStack(stack *S){
	(*S) = (stack)malloc(sizeof(_stack));                           
	(*S)->base = (point*) malloc ( _size*sizeof(_point) );
	(*S)->top = (*S)->base;
	(*S)->size = _size;
}

void push(stack st,point p){
	st->top++;
	*st->top = p;
}

point pop(stack st){
	point p = *st->top;
	st->top--;
	return p;
}

int length(stack st){
	return st->top - st->base;
}

int isEmpty(stack st){
	if(length(st) == 0)
		return 1;
	return 0;
}

point getPoint(int i,int j){
	point p = (point)malloc(sizeof(_point));
	p->x = i;
	p->y = j;
	return p;
}


int place(_stack s,int x,int y){
	stack st = &s;
	point p;
	while(!isEmpty(st)){
		p = pop(st);
		if(p->x == x || p->y == y || abs(x - p->x) == abs(y - p->y) )
			return 0;
	}
		
	return 1;
}

int main(){

	int i,j;
	stack st;
	point p;
	InitStack(&st);
	push(st,getPoint(0,0));
	i=1,j=0;

	do{
		for(;i<len;i++){

			for(;j<len;j++)
				if(place(*st,i,j)){
					push(st,getPoint(i,j));
					if(length(st) == len)
						goto endF;
				}

			j = 0;
		}

	p = pop(st);
	i = p->x;
	j = p->y + 1;
	}while(!isEmpty(st));

endF:
	while(!isEmpty(st)){
		p = pop(st);
		printf("(%d,%d) => ",p->x + 1,p->y + 1);
	}
	printf("\n");
}