leetcode N-Queens/N-Queens II, backtracking, hdu 2553 count N-Queens, dfs 分类： leetcode hdoj 2015-07-09 02:07 102人阅读 评论(0) 收藏

1. 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.

2. 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;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。// p.s. If in any way improment can be achieved, better performance or whatever, it will be well-appreciated to let me know, thanks in advance.