相关问题1:[LeetCode] N-Queen N皇后
相关问题2:
八皇后问题,是一个古老而著名的问题,是回溯算法的典型例题。
这里给出八皇后问题的解答。思路是:对解答树进行深度遍历,当遍历抵达第8层的时候,我们便找到了一个解。如果在抵达第8层之前,发现解答树的节点违反了八皇后问题的要求,那么进行回溯。
具体的代码如下:
#include<iostream>
#include<vector>
#include<iterator>
using namespace std;
bool test_col(int c, vector<bool> col)
{// true if the c-th column is occupied.
return col.at(c);
}
bool test_main(int r, int c, vector<bool> main_diag)
{// true if the main diag passing through (r,c) is occupied
return main_diag.at(r-c+7);
}
bool test_counter(int r, int c, vector<bool> counter_diag)
{// true if the counter diag passing through (r, c) is occupied
return counter_diag.at(r+c);
}
void eight_queen(vector<int> pos, vector<bool> col, vector<bool> main_diag, vector<bool> counter_diag)
{
if(pos.size()==8)
{
// copy can be used to print out the elements in an iterator range
copy(pos.begin(), pos.end(), ostream_iterator<int>(cout, "|") );
cout<<"\n";
return;
}
else
{
for(int i=0; i<8; i++)
{
if(!(test_col(i, col) || test_main(pos.size(), i, main_diag) || test_counter(pos.size(), i, counter_diag) ))
{
col.at(i) = true;
main_diag.at(pos.size()-i+7) = true;
counter_diag.at(pos.size()+i) = true;
pos.push_back(i);
eight_queen(pos, col, main_diag, counter_diag);
pos.pop_back();
col.at(i) = 0;
main_diag.at(pos.size()-i+7) = 0;
counter_diag.at(pos.size()+i) = 0;
}
}
}
return;
}
int main()
{
vector<bool> col(8, 0); // false: no queen is occupying this column. true: a queen is now occupying this column.
vector<bool> main_diag(15, 0); // false: no queen is occupying this main diag. true: a queen is now occupying this main diag.
vector<bool> counter_diag(15, 0); // false: no queen is occupying this counter diag. true: a queen is now occupying this counter diag.
vector<int> pos;
eight_queen(pos, col, main_diag, counter_diag);
}