n皇后问题-回溯法

用树可以形象直观地描述回溯法。《算法设计技巧与分析》

#ifndef __QUEENS_H__#define __QUEENS_H__class Queens{public:Queens(int nQueens);~Queens();bool Setout(); // 摆放皇后inline int* GetAnswer(){return m_pos;}inline bool isLegal(int row, int col) // 合法性{return (row == m_nQueens-1) && isPartSolution(row, col);}inline bool isPartSolution(int row, int col) //在第row行的第col列摆放{// 判断当前的位置与原先排好的位置没有冲突int i, j;bool flag = true;for (i = 0; i < row; i++){j = m_pos[i];if (col == j || row-i == col-j || row-i == j-col){flag = false;break;}}return flag;}protected:int m_nQueens; // 皇后数目int* m_pos; };#endif

#include "queens.h"#include <assert.h>#include <string.h>// constructorQueens::Queens(int nQueens){m_nQueens = nQueens;m_pos = new int[m_nQueens];assert(m_pos);}// destructorQueens::~Queens(){if (m_pos != 0){delete[] m_pos;}}// 摆放皇后bool Queens::Setout(){// 设置所有位置为空，不摆放皇后/*memset(m_pos, 0, sizeof(int) * m_nQueens);*/int i = 0;for (i = 0; i < m_nQueens; i++){m_pos[i] = -1;}bool flag = false; // 标记是否找到合法的解int k = 0; // 第k行while (k >= 0){while (m_pos[k] < m_nQueens-1) // m_pos[k]的最大值为m_nQueens-1{// 当m_pos[k]取到最大值后，仍然没有得到合法解时，跳出内循环进行回溯m_pos[k] += 1;if (isLegal(k, m_pos[k])){flag = true;break;}else if (isPartSolution(k, m_pos[k])){k++; }}if (flag){break;}else{// 回溯m_pos[k] = -1; // 清楚当前层k--;}}return flag;}

#include <iostream>#include "queens.h"using namespace std;int main(){int const nQueens = 10;Queens queens(nQueens);if (queens.Setout()){int* answer = queens.GetAnswer();int i, j;for (i = 0; i<nQueens; i++){for (j = 0; j < nQueens; j++){if (j == answer[i]){cout << "+" << " ";}else{cout << "-" << " ";}}cout << endl;}}else{cout << "no solution" << endl;}return 0;}