Leetcode: N-Queens II

N-Queen ||

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.



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N皇后不打印版本。

 回溯：以行为基础，row[ i ] 表示 第 i 行的 queen 所在的列号， cal[ i ] 表示 第 i 列是否有皇后 ，判断 对角线 abs(j-r)==abs(row[j]-row[r]) ？1:0 遍历每一种排列情况就可以得到答案

#include<vector>
#include<iostream>
#include<cstdio>
#include<math.h>
#include<string.h>
#include<algorithm>
using namespace std;

class Solution {
public:
int answer;
void solve(int r,int n,int * row,int * cal ){
if(r>n){
answer++;
return;
}
for(int i=1;i<=n;i++){
if(!cal[i]){
cal[i]=1;
row[r]=i;
for(int j=1;j<r;j++){
if(abs(j-r)==abs(row[j]-row[r])){
row[r]=0;
cal[i]=0;
}
}
if(cal[i] && row[r]){
solve(r+1,n,row,cal);
}
row[r]=0;
cal[i]=0;

}
}
}
int totalNQueens(int n) {
answer=0;
int row[n+1];
int cal[n+1];
memset(row,0,sizeof(row));
memset(cal,0,sizeof(cal));
solve(1,n,row,cal);
return answer;
}
};
int main()
{
int answer;
Solution A;
answer=A.totalNQueens(8);
cout<<answer<<endl;

}