每次需要满足的条件：abs(x[j]-x[k])==abs(j-k) || x[j]==x[k]

#include <iostream>
#include <cmath>
using namespace std;
const int num=8;
int sum=0;
int x[num]={0};
bool place (int k)
{
    for(int j=1;j<k;j++)
    {
        if(abs(x[j]-x[k])==abs(j-k)||x[j]==x[k])
            return false;
    }
    return true;
}
void backtrack(int t)
{
    if(t>num)
    {
        sum++;
        for(int m=1;m<=num;m++)
        {
            cout<<x[m];
        }
        cout<<endl;
    }
    else
    {
        for(int i=1;i<=num;i++)
        {
            x[t]=i;
            if(place(t))backtrack(t+1);
        }
    }
}

int main()
{
    backtrack(1);
    cout<<"All solutions: "<<sum;
    return 0;
}

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