2014-03-20 04:08
题目:八皇后问题。
解法:DFS解决。
代码:
1 // 9.9 Eight-Queen Problem, need I say more? 2 #include <cstdio> 3 #include <vector> 4 using namespace std; 5 6 class Solution 7 { 8 public: 9 void eigthQueens(int n, vector<vector<int> > &result) { 10 if (n < 1) { 11 return; 12 } 13 14 vector<int> v; 15 dfs(0, n, v, result); 16 v.clear(); 17 } 18 private: 19 void dfs(int idx, int n, vector<int> &v, vector<vector<int> > &result) { 20 if (idx == n) { 21 result.push_back(v); 22 return; 23 } 24 25 int i, j; 26 for (i = 0; i < n; ++i) { 27 for (j = 0; j < idx; ++j) { 28 if (v[j] == i || myabs(idx - j) == myabs(i - v[j])) { 29 // conflict 30 break; 31 } 32 } 33 if (j == idx) { 34 v.push_back(i); 35 dfs(idx + 1, n, v, result); 36 v.pop_back(); 37 } 38 } 39 } 40 41 int myabs(int n) { 42 return (n >= 0 ? n : -n); 43 } 44 }; 45 46 int main() 47 { 48 vector<vector<int> > result; 49 Solution sol; 50 int n; 51 int i, j, k; 52 53 while (scanf("%d", &n) == 1 && n > 0) { 54 sol.eigthQueens(n, result); 55 for (k = 0; k < (int)result.size(); ++k) { 56 for (i = 0; i < n; ++i) { 57 for (j = 0; j < result[k][i]; ++j) { 58 putchar('0'); 59 } 60 putchar('1'); 61 for (j = result[k][i] + 1; j < n; ++j) { 62 putchar('0'); 63 } 64 putchar(' '); 65 } 66 putchar(' '); 67 } 68 printf("%d ", (int)result.size()); 69 for (i = 0; i < (int)result.size(); ++i) { 70 result[i].clear(); 71 } 72 result.clear(); 73 } 74 75 return 0; 76 }