摘要: N 后问题说明, 一山不容二只母老虎.
1. 代码先上代码, 再说废话.
#include2. 运行结果#include #include bool place(int* paraSolution, int paraT){ int j; for (j = 1; j < paraT; j ++){ if ((abs(paraT - j) == abs(paraSolution[j] - paraSolution[paraT])) || (paraSolution[j] == paraSolution[paraT])) return false; }//Of for return true; }//Of place void backtracking(int* paraSolution, int paraN, int paraT){ int i; if (paraT > paraN){ for (i = 1; i <= paraN; i ++) printf("%d ", paraSolution[i]); printf("rn"); }else{ for (i = 1; i <= paraN; i ++){ paraSolution[paraT] = i; if (place(paraSolution, paraT)) backtracking(paraSolution, paraN, paraT + 1); }//Of for i }//Of if }//Of backtracking void nQueen(int paraN){ int i; int* solution = (int*)malloc((paraN + 1) * sizeof(int)); for (i = 0; i <= paraN; i ++) solution[i] = 0; backtracking(solution, paraN, 1); }//Of nQueen int main(){ nQueen(5); return 1; }//Of main
1 3 5 2 4 1 4 2 5 3 2 4 1 3 5 2 5 3 1 4 3 1 4 2 5 3 5 2 4 1 4 1 3 5 2 4 2 5 3 1 5 2 4 1 3 5 3 1 4 2 Press any key to continue3. 代码分析
- 回溯算法是万能解题法.
- 回溯法的框架都是一致的.



