leetcode N-Queens/N-Queens II, backtracking, hdu 2553 count N-Queens, dfs ...-优快云博客

本文深入探讨了N皇后问题的解决策略，通过改进的回溯算法和深度优先搜索优化，实现高效的解决方案。分析了验证检查过程中的优化策略，包括额外的占用数组来提高效率，并提供了具体的实现代码，包括简化版和完整版解决方案。

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for the backtracking part, thanks to the video of stanford cs106b lecture 10 by Julie Zelenski for the nice explanation of recursion and backtracking, highly recommended.
in hdu 2553 cout N-Queens solutions problem, dfs is used.
// please ignore, below analysis is inaccurate, though for inspiration considerations, I decided to let it remain.
and we use a extra occupied[] array to opimize the validation test, early termination when occupied[i]==1, and in function isSafe the test change from 3 to 2 comparisons.
for the improvement, here is an analysis:
noted that total call of dfs for n is less than factorial(n), for convenience we take as factorial(n),
in this appoach, in pow(n,n)-factorial(n) cases, we conduct one comparison, and three for others
otherwise, for all pow(n,n) cases, we must conduct three comparisons,
blow is a list for n from 1 to 10,

   n        n!           n^n   n!/n^n
   1        1             1   1.00000
   2        2             4   0.50000
   3        6            27   0.22222
   4       24           256   0.09375
   5      120          3125   0.03840
   6      720         46656   0.01543
   7     5040        823543   0.00612
   8    40320      16777216   0.00240
   9   362880     387420489   0.00094
  10  3628800   10000000000   0.00036

the table shows that, for almost all of the cases of large n (n>=4), we reduce 3 to 1 comparison in doing validation check.

occupied[i] is initilized to 0, before dfs, mark.

occupied[val]=1;

after dfs, unmark,

occupied[val]=0;

where val is the value chosen.
inspirations from chapter 3 Decompositions of graphs
in Algorithms(算法概论), Sanjoy Dasgupta University of California, San Diego Christos Papadimitriou University of California at Berkeley Umesh Vazirani University of California at Berkeley.

// leetcode N-Queens(12ms)

class Solution {
    vector<vector<string>> ans;
    int N;
    //int numofsol;

    void AddTo_ans(string &conf) {
        vector<string> vecstrtmp;
        string strtmp(N,'.');
        for(auto ind: conf) {
            vecstrtmp.push_back(strtmp);
            vecstrtmp.back()[(size_t)ind-1]='Q';
        }
        ans.push_back(vector<string>());
        ans.back().swap(vecstrtmp);
    }
    void RecListNQueens(string soFar, string rest) {
        if(rest.empty()) {
            //++numofsol;
            AddTo_ans(soFar);
        }
        else {
            int i,j,len=soFar.size(), flag;
            for(i=0;i<rest.size();++i) {
                for(j=0;j<len;++j) {
                    flag=1;
                    if(soFar[j]-rest[i]==len-j || soFar[j]-rest[i]==j-len) {
                        flag=0; break;
                    }
                }
                if(flag) RecListNQueens(soFar+rest[i],rest.substr(0,i)+rest.substr(i+1));
            }
        }
    }
public:
    vector<vector<string>> solveNQueens(int n) {
        ans.clear();
        //numofsol=0;
        N=n;
        string str;
        for(int i=1;i<=n;++i) str.push_back(i);
        RecListNQueens("", str);
        return ans;
    }
};

// with a tiny modification,
// leetcode N-Queens II (8ms)

class Solution {
    //vector<vector<string>> ans;
    int N;
    int numofsol;

    /*void AddTo_ans(string &conf) {
        vector<string> vecstrtmp;
        string strtmp(N,'.');
        for(auto ind: conf) {
            vecstrtmp.push_back(strtmp);
            vecstrtmp.back()[(size_t)ind-1]='Q';
        }
        ans.push_back(vector<string>());
        ans.back().swap(vecstrtmp);
    }*/
    void RecListNQueens(string soFar, string rest) {
        if(rest.empty()) {
            ++numofsol;
            //AddTo_ans(soFar);
        }
        else {
            int i,j,len=soFar.size(), flag;
            for(i=0;i<rest.size();++i) {
                for(j=0;j<len;++j) {
                    flag=1;
                    if(soFar[j]-rest[i]==len-j || soFar[j]-rest[i]==j-len) {
                        flag=0; break;
                    }
                }
                if(flag) RecListNQueens(soFar+rest[i],rest.substr(0,i)+rest.substr(i+1));
            }
        }
    }
public:
    int totalNQueens(int n) {
        //ans.clear();
        numofsol=0;
        N=n;
        string str;
        for(int i=1;i<=n;++i) str.push_back(i);
        RecListNQueens("", str);
        return numofsol;
    }
};

// hdu 2553, 15ms

#include <cstdio>
#include <vector>
#include <algorithm>

class countNQueenSol {
    static const int MAXN=12;
    static int values[MAXN];
    static int occupied[MAXN];
    static std::vector<int> ans;
    static int cnt, rownum;

    static bool isSafe(int val, int row) {
        for(int i=0;i<row;++i) {
            if(val-values[i]==row-i || val-values[i]==i-row)
            return false;
        }
        return true;
    }
    static void dfs(int row) {
        if(row==rownum) ++cnt;
        for(int i=0;i<rownum;++i) {
            if(occupied[i]==0 && isSafe(i,row)) {
                values[row]=i;
                occupied[i]=1;
                dfs(row+1);
                occupied[i]=0;
            }
        }
    }
public:
    static int getMAXN() { return MAXN; }
    static int solve(int k) {
        if(k<=0 || k>=countNQueenSol::MAXN) return -1;
        if(ans[k]<0) {
            cnt=0;
            rownum=k;
            dfs(0);
            //if(k==0) cnt=0; // adjust anomaly when k==0
            ans[k]=cnt;
        }
        return ans[k];
    }
};
int countNQueenSol::values[countNQueenSol::MAXN]={0};
int countNQueenSol::occupied[countNQueenSol::MAXN]={0};
std::vector<int> countNQueenSol::ans(countNQueenSol::MAXN,-1);
int countNQueenSol::cnt, countNQueenSol::rownum;

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif
    int n;
    while(scanf("%d",&n)==1 && n>0) {
        printf("%d\n",countNQueenSol::solve(n));
    }
    return 0;
}