《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 < ) {
             return;
         }

         vector<int> v;
         dfs(, 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 = ; i < n; ++i) {
             for (j = ; 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 + , n, v, result);
                 v.pop_back();
             }
         }
     }

     int myabs(int n) {
         return (n >=  ? n : -n);
     }
 };

 int main()
 {
     vector<vector<int> > result;
     Solution sol;
     int n;
     int i, j, k;

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

     return ;
 }