《Cracking the Coding Interview》——第9章：递归和动态规划——题目9

2014-03-20 04:08

题目：八皇后问题。

解法：DFS解决。

代码：

// 9.9 Eight-Queen Problem, need I say more?
#include <cstdio>
#include <vector>
using namespace std;

class Solution
{
public:
    void eigthQueens(int n, vector<vector<int> > &result) {
        if (n < 1) {
            return;
        }
        
        vector<int> v;
        dfs(0, n, v, result);
        v.clear();
    }
private:
    void dfs(int idx, int n, vector<int> &v, vector<vector<int> > &result) {
        if (idx == n) {
            result.push_back(v);
            return;
        }

        int i, j;
        for (i = 0; i < n; ++i) {
            for (j = 0; j < idx; ++j) {
                if (v[j] == i || myabs(idx - j) == myabs(i - v[j])) {
                    // conflict
                    break;
                }
            }
            if (j == idx) {
                v.push_back(i);
                dfs(idx + 1, n, v, result);
                v.pop_back();
            }
        }
    }
    
    int myabs(int n) {    
        return (n >= 0 ? n : -n);
    }
};

int main()
{
    vector<vector<int> > result;
    Solution sol;
    int n;
    int i, j, k;
    
    while (scanf("%d", &n) == 1 && n > 0) {
        sol.eigthQueens(n, result);
        for (k = 0; k < (int)result.size(); ++k) {
            for (i = 0; i < n; ++i) {
                for (j = 0; j < result[k][i]; ++j) {
                    putchar('0');
                }
                putchar('1');
                for (j = result[k][i] + 1; j < n; ++j) {
                    putchar('0');
                }
                putchar('\n');
            }
            putchar('\n');
        }
        printf("%d\n", (int)result.size());
        for (i = 0; i < (int)result.size(); ++i) {
            result[i].clear();
        }
        result.clear();
    }
    
    return 0;
}