八皇後問題c語言遞歸

⑴ 求教C語言回溯法寫出八皇後問題的92種解

(1)全排列

將自然數1~n進行排列，共形成n!中排列方式，叫做全排列。

例如3的全排列是：1/2/3、1/3/2、2/1/3、2/3/1、3/1/2、3/2/1，共3!=6種。

(2)8皇後（或者n皇後）

保證8個皇後不能互相攻擊，即保證每一橫行、每一豎行、每一斜行最多一個皇後。

我們撇開第三個條件，如果每一橫行、每一豎行都只有一個皇後。

將8*8棋盤標上坐標。我們討論其中的一種解法：

- - - - - - - Q

- - - Q - - - -

Q - - - - - - -

- - Q - - - - -

- - - - - Q - -

- Q - - - - - -

- - - - - - Q -

- - - - Q - - -

如果用坐標表示就是：(1,8) (2,4) (3,1) (4,3) (5,6) (6,2) (7,7) (8,5)

將橫坐標按次序排列，縱坐標就是8/4/1/3/6/2/7/5。這就是1~8的一個全排列。

我們將1~8的全排列存入輸入a[]中(a[0]~a[7])，然後8個皇後的坐標就是(i+1,a[i])，其中i為0~7。

這樣就能保證任意兩個不會同一行、同一列了。

置於斜行，你知道的，兩個點之間連線的斜率絕對值為1或者-1即為同一斜行，充要條件是|x1-x2|=|y1-y2|（兩個點的坐標為（x1,y1）(x2,y2)）。我們在輸出的時候進行判斷，任意兩個點如果滿足上述等式，則判為失敗，不輸出。

下面附上代碼：添加必要的注釋，其中全排列的實現看看注釋應該可以看懂：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
intprinted;
//該函數用於畫圖，這里為了節約空間則略去
//讀者只需要將draw(a,k);去掉注釋即可畫圖
voiddraw(int*a,intk)
{
	inti,j;
	for(i=0;i<k;i++)
	{
		printf("	");
		for(j=0;j<k;j++)
			//有皇後輸出Q，否則輸出-
			if(a[i]-1==j)printf("Q");elseprintf("-");
		printf("
");
	}
	printf("
");

}
//遞歸實現全排列,a是數組，iStep是位置的測試點，k是皇後的個數，一般等於8
voidSettle(int*a,intiStep,intk)
{	
	inti,j,l,flag=1;
	//如果iStep的數字等於a之前的數字，則存在重復，返回
	for(i=0;i<iStep-1;i++)
		if(a[iStep-1]==a[i])return;
	//如果iStep==k，即遞歸結束到最後一位，可以驗證是否斜行滿足
	if(iStep==k)
	{
		//雙重循環判斷是否斜行滿足
		for(j=0;j<k;j++)
			for(l=0;l<k&&l!=j;l++)
				//如果不滿足，則flag=0
				if(fabs(j-l)==fabs(a[j]-a[l]))flag=0;
		//如果flag==1，則通過了斜行的所有測試，輸出。
		if(flag)
		{
			for(i=0;i<k;i++)
				printf("(%d,%d)",i+1,a[i]);
			printf("
");
			//如果去掉這里的注釋可以獲得畫圖，由於空間不夠，這里略去
		//	draw(a,k);
			//printed變數計算有多少滿足題意的結果，是全局變數
			printed++;
		}
		flag=1;
	}
	//如果未測試至最後末尾，則測試下一位（遞歸）
	for(i=1;i<=k;i++)
	{
		a[iStep]=i;
		Settle(a,iStep+1,k);
	}
}
voidmain()
{
	int*a;
	intk;
	//輸入維數，建立數組
	printf("Enterthesizeofthesquare:");
	scanf("%d",&k);
	a=(int*)calloc(k,sizeof(int));
	//清屏，從iStep=0處進入遞歸
	system("cls");
	Settle(a,0,k);
	//判斷最後是否有結果
	if(!printed)printf("Noanswersaccepted!
");
	elseprintf("%dstatesavailable!
",printed);
}
附輸出結果（輸入k=8）：
(1,1)(2,5)(3,8)(4,6)(5,3)(6,7)(7,2)(8,4)
(1,1)(2,6)(3,8)(4,3)(5,7)(6,4)(7,2)(8,5)
(1,1)(2,7)(3,4)(4,6)(5,8)(6,2)(7,5)(8,3)
(1,1)(2,7)(3,5)(4,8)(5,2)(6,4)(7,6)(8,3)
(1,2)(2,4)(3,6)(4,8)(5,3)(6,1)(7,7)(8,5)
(1,2)(2,5)(3,7)(4,1)(5,3)(6,8)(7,6)(8,4)
(1,2)(2,5)(3,7)(4,4)(5,1)(6,8)(7,6)(8,3)
(1,2)(2,6)(3,1)(4,7)(5,4)(6,8)(7,3)(8,5)
(1,2)(2,6)(3,8)(4,3)(5,1)(6,4)(7,7)(8,5)
(1,2)(2,7)(3,3)(4,6)(5,8)(6,5)(7,1)(8,4)
(1,2)(2,7)(3,5)(4,8)(5,1)(6,4)(7,6)(8,3)
(1,2)(2,8)(3,6)(4,1)(5,3)(6,5)(7,7)(8,4)
(1,3)(2,1)(3,7)(4,5)(5,8)(6,2)(7,4)(8,6)
(1,3)(2,5)(3,2)(4,8)(5,1)(6,7)(7,4)(8,6)
(1,3)(2,5)(3,2)(4,8)(5,6)(6,4)(7,7)(8,1)
(1,3)(2,5)(3,7)(4,1)(5,4)(6,2)(7,8)(8,6)
(1,3)(2,5)(3,8)(4,4)(5,1)(6,7)(7,2)(8,6)
(1,3)(2,6)(3,2)(4,5)(5,8)(6,1)(7,7)(8,4)
(1,3)(2,6)(3,2)(4,7)(5,1)(6,4)(7,8)(8,5)
(1,3)(2,6)(3,2)(4,7)(5,5)(6,1)(7,8)(8,4)
(1,3)(2,6)(3,4)(4,1)(5,8)(6,5)(7,7)(8,2)
(1,3)(2,6)(3,4)(4,2)(5,8)(6,5)(7,7)(8,1)
(1,3)(2,6)(3,8)(4,1)(5,4)(6,7)(7,5)(8,2)
(1,3)(2,6)(3,8)(4,1)(5,5)(6,7)(7,2)(8,4)
(1,3)(2,6)(3,8)(4,2)(5,4)(6,1)(7,7)(8,5)
(1,3)(2,7)(3,2)(4,8)(5,5)(6,1)(7,4)(8,6)
(1,3)(2,7)(3,2)(4,8)(5,6)(6,4)(7,1)(8,5)
(1,3)(2,8)(3,4)(4,7)(5,1)(6,6)(7,2)(8,5)
(1,4)(2,1)(3,5)(4,8)(5,2)(6,7)(7,3)(8,6)
(1,4)(2,1)(3,5)(4,8)(5,6)(6,3)(7,7)(8,2)
(1,4)(2,2)(3,5)(4,8)(5,6)(6,1)(7,3)(8,7)
(1,4)(2,2)(3,7)(4,3)(5,6)(6,8)(7,1)(8,5)
(1,4)(2,2)(3,7)(4,3)(5,6)(6,8)(7,5)(8,1)
(1,4)(2,2)(3,7)(4,5)(5,1)(6,8)(7,6)(8,3)
(1,4)(2,2)(3,8)(4,5)(5,7)(6,1)(7,3)(8,6)
(1,4)(2,2)(3,8)(4,6)(5,1)(6,3)(7,5)(8,7)
(1,4)(2,6)(3,1)(4,5)(5,2)(6,8)(7,3)(8,7)
(1,4)(2,6)(3,8)(4,2)(5,7)(6,1)(7,3)(8,5)
(1,4)(2,6)(3,8)(4,3)(5,1)(6,7)(7,5)(8,2)
(1,4)(2,7)(3,1)(4,8)(5,5)(6,2)(7,6)(8,3)
(1,4)(2,7)(3,3)(4,8)(5,2)(6,5)(7,1)(8,6)
(1,4)(2,7)(3,5)(4,2)(5,6)(6,1)(7,3)(8,8)
(1,4)(2,7)(3,5)(4,3)(5,1)(6,6)(7,8)(8,2)
(1,4)(2,8)(3,1)(4,3)(5,6)(6,2)(7,7)(8,5)
(1,4)(2,8)(3,1)(4,5)(5,7)(6,2)(7,6)(8,3)
(1,4)(2,8)(3,5)(4,3)(5,1)(6,7)(7,2)(8,6)
(1,5)(2,1)(3,4)(4,6)(5,8)(6,2)(7,7)(8,3)
(1,5)(2,1)(3,8)(4,4)(5,2)(6,7)(7,3)(8,6)
(1,5)(2,1)(3,8)(4,6)(5,3)(6,7)(7,2)(8,4)
(1,5)(2,2)(3,4)(4,6)(5,8)(6,3)(7,1)(8,7)
(1,5)(2,2)(3,4)(4,7)(5,3)(6,8)(7,6)(8,1)
(1,5)(2,2)(3,6)(4,1)(5,7)(6,4)(7,8)(8,3)
(1,5)(2,2)(3,8)(4,1)(5,4)(6,7)(7,3)(8,6)
(1,5)(2,3)(3,1)(4,6)(5,8)(6,2)(7,4)(8,7)
(1,5)(2,3)(3,1)(4,7)(5,2)(6,8)(7,6)(8,4)
(1,5)(2,3)(3,8)(4,4)(5,7)(6,1)(7,6)(8,2)
(1,5)(2,7)(3,1)(4,3)(5,8)(6,6)(7,4)(8,2)
(1,5)(2,7)(3,1)(4,4)(5,2)(6,8)(7,6)(8,3)
(1,5)(2,7)(3,2)(4,4)(5,8)(6,1)(7,3)(8,6)
(1,5)(2,7)(3,2)(4,6)(5,3)(6,1)(7,4)(8,8)
(1,5)(2,7)(3,2)(4,6)(5,3)(6,1)(7,8)(8,4)
(1,5)(2,7)(3,4)(4,1)(5,3)(6,8)(7,6)(8,2)
(1,5)(2,8)(3,4)(4,1)(5,3)(6,6)(7,2)(8,7)
(1,5)(2,8)(3,4)(4,1)(5,7)(6,2)(7,6)(8,3)
(1,6)(2,1)(3,5)(4,2)(5,8)(6,3)(7,7)(8,4)
(1,6)(2,2)(3,7)(4,1)(5,3)(6,5)(7,8)(8,4)
(1,6)(2,2)(3,7)(4,1)(5,4)(6,8)(7,5)(8,3)
(1,6)(2,3)(3,1)(4,7)(5,5)(6,8)(7,2)(8,4)
(1,6)(2,3)(3,1)(4,8)(5,4)(6,2)(7,7)(8,5)
(1,6)(2,3)(3,1)(4,8)(5,5)(6,2)(7,4)(8,7)
(1,6)(2,3)(3,5)(4,7)(5,1)(6,4)(7,2)(8,8)
(1,6)(2,3)(3,5)(4,8)(5,1)(6,4)(7,2)(8,7)
(1,6)(2,3)(3,7)(4,2)(5,4)(6,8)(7,1)(8,5)
(1,6)(2,3)(3,7)(4,2)(5,8)(6,5)(7,1)(8,4)
(1,6)(2,3)(3,7)(4,4)(5,1)(6,8)(7,2)(8,5)
(1,6)(2,4)(3,1)(4,5)(5,8)(6,2)(7,7)(8,3)
(1,6)(2,4)(3,2)(4,8)(5,5)(6,7)(7,1)(8,3)
(1,6)(2,4)(3,7)(4,1)(5,3)(6,5)(7,2)(8,8)
(1,6)(2,4)(3,7)(4,1)(5,8)(6,2)(7,5)(8,3)
(1,6)(2,8)(3,2)(4,4)(5,1)(6,7)(7,5)(8,3)
(1,7)(2,1)(3,3)(4,8)(5,6)(6,4)(7,2)(8,5)
(1,7)(2,2)(3,4)(4,1)(5,8)(6,5)(7,3)(8,6)
(1,7)(2,2)(3,6)(4,3)(5,1)(6,4)(7,8)(8,5)
(1,7)(2,3)(3,1)(4,6)(5,8)(6,5)(7,2)(8,4)
(1,7)(2,3)(3,8)(4,2)(5,5)(6,1)(7,6)(8,4)
(1,7)(2,4)(3,2)(4,5)(5,8)(6,1)(7,3)(8,6)
(1,7)(2,4)(3,2)(4,8)(5,6)(6,1)(7,3)(8,5)
(1,7)(2,5)(3,3)(4,1)(5,6)(6,8)(7,2)(8,4)
(1,8)(2,2)(3,4)(4,1)(5,7)(6,5)(7,3)(8,6)
(1,8)(2,2)(3,5)(4,3)(5,1)(6,7)(7,4)(8,6)
(1,8)(2,3)(3,1)(4,6)(5,2)(6,5)(7,7)(8,4)
(1,8)(2,4)(3,1)(4,3)(5,6)(6,2)(7,7)(8,5)
92statesavailable!

⑵ 八皇後c++求解思路及代碼（遞歸、回溯）

八皇後問題的解法，採用遞歸與回溯演算法。總解數為92種，但歸納後僅42類，考慮旋轉與對稱情況，歸於一類。此C++代碼簡潔明了，全面呈現所有可能。

解題核心在於，通過窮舉評估下一個皇後位置與前皇後位置是否有沖突，若有則回溯修正位置。當當前位置皇後安置完成後，遞歸繼續探索下一個皇後位置。

遞歸法具高可讀性，但循環法亦可實現。八皇後問題的C++代碼及92種求解方案已整理完畢。