b)若当前行不是最后一行,当前行设为下一行, 当前列设为当前行的第一个待测位置; c)若当前行是最后一行,当前列不是最后一列,当前列设为下一列; d)若当前行是最后一行,当前列是最后一列,回溯,即清空当前行及以下各行的棋盘,然后,当前行设为上一行,当前列设为当前行的下一个待测位置。 e)以上返回到第2...
b)若当前行不是最后一行,当前行设为下一行, 当前列设为当前行的第一个待测位置; c)若当前行是最后一行,当前列不是最后一列,当前列设为下一列; d)若当前行是最后一行,当前列是最后一列,回溯,即清空当前行及以下各行的棋盘,然后,当前行设为上一行,当前列设为当前行的下一个待测位置。 e)以上返回到第2...
今早起来写了个N皇后 #include <iostream>#defineabs(i) ((i) < 0 ? -(i) : (i))constintN =8;intboard[N] = {-1};boolok(introw) {if(row ==0)returntrue;for(inti =0; i < row; ++i) {if(((row - i) == abs(board[row] - board[i])) ||board[row]- board[i] ==0)r...
b)若当前行不是最后一行,当前行设为下一行, 当前列设为当前行的第一个待测位置; c)若当前行是最后一行,当前列不是最后一列,当前列设为下一列; d)若当前行是最后一行,当前列是最后一列,回溯,即清空当前行及以下各行的棋盘,然后,当前行设为上一行,当前列设为当前行的下一个待测位置。 e)以上返回到第2...
//version 1回溯#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#defineMAXN 100010usingnamespacestd;boolhassolve=false;//是否解出intn;intline[MAXN];boolvis[3][MAXN];voidsolve(intlayer)//layer表示的是已经完成的最高层数{if(hassolve==true)return;if(layer==n) ...
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> const int N=1e5+5; int n; int len; char s[20]; int t; int x[20];//x数组表示,第i行(即x数组的下标),x[i]存储的是第i行的下标 int jieguo[20]; bool yes(int k) { for(int i=0;i<k;i++) { if(...
直接规约成SAT问题，然后SAT问题的解法一大把dancing links algorithm还可以用来解数独本文将介绍N皇后问题...
/// main.cpp// BackTrack Solution of N-Queens Problem./// Created by Kang on 2020/7/2 at NJUPT.// Copyright © 2020 Kang. All rights reserved.//#include<iostream>#include<cmath>#include<ctime>using namespace std;constint maxSize=10;int x[maxSize];/** Judge...
In 1979 Hitotumatu and Noshita gave an algorithm for producing solutions to the N queens problem in lexicographical order based directly on the generation of permutations. They were not the first to base an algorithm on the use of permutations, but they did produce the solutions in ...
Here you will get program for N queens problem in C using backtracking. N Queens Problem is a famous puzzle in which n-queens are to be placed on a nxn chess board such that no two queens are in the same row, column or diagonal.