最近在学习Lua脚本，经过了不到十天的学习，也算是对语法有所了解吧，，另外正好也看到了八皇后问题，感觉挺有意思的 就试了试用算法解出来。
八皇后问题的原题是：八皇后问题是一个以国际象棋为背景的问题：如何能够在 8×8 的国际象棋棋盘上放置八个皇后，使得任何一个皇后都无法直接吃掉其他的皇后？为了达到此目的，任两个皇后都不能处于同一条横行、纵行或斜线上。
以下是lua的算法代码：
local eightQueen = { 0,0,0,0,0,0,0,0,}local count = 0function check(row,column)–检查local datafor i = 1,row dodata = eightQueen[i]if column == data then–同列return falseelseif (i + data) == (row + column) then–/斜线return falseelseif (i – data) == (row – column) then–\斜线return falseendendreturn trueendfunction eight_queen(row)for column = 1,8 doif check(row,column) == true theneightQueen[row] = columnif row == 8 thencount = count + 1eightQueen[row] = 0returnendeight_queen(row + 1)eightQueen[row] = 0endendendeight_queen(1)print(count)思路照抄百度百科的C++代码 C++代码如下：
#include<iostream>using namespace std;static int gEightQueen[8] = { 0 }, gCount = 0;void print()//输出每一种情况下棋盘中皇后的摆放情况{for (int outer = 0; outer < 8; outer++){for (int inner = 0; inner < gEightQueen[outer]; inner++)cout << "#";for (int inner = gEightQueen[outer] + 1; inner < 8; inner++)cout << "";cout << endl;}cout << "==========================\n";}int check_pos_valid(int loop, int value)//检查是否存在有多个皇后在同一行/列/对角线的情况{int index;int data;for (index = 0; index < loop; index++){data = gEightQueen[index];if (value == data)return 0;if ((index + data) == (loop + value))return 0;if ((index – data) == (loop – value))return 0;}return 1;}void eight_queen(int index){int loop;for (loop = 0; loop < 8; loop++){if (check_pos_valid(index, loop)){gEightQueen[index] = loop;if (7 == index){gCount++, print();gEightQueen[index] = 0;return;}eight_queen(index + 1);gEightQueen[index] = 0;}}}int main(int argc, char*argv[]){eight_queen(0);cout << "total=" << gCount << endl;return 0;}输出如下：
己欲立先立人，已欲达先达人。
