多维容器按列组合元素

<pre name="code" class="cpp">// alg2.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"//昨天碰到一个文件，有这么一个std::vector<std::vector<std::string>> //现在需要将每一列的数据进行组合//按照普通写法，每个vector的长度是知道的直接//for(int lvl1;.....)// for(int lvl2;...)//for(int lvl3,...)//但问题是第一层的长度是变长，所以无法按照vector写出。//木有办法啊，代码还是要写，还是要将结果组合出来//通过递归来完成操作。//made by davidsu33 2014-6-25#include <boost/config/warning_disable.hpp>#include <vector>#include <string>#include <iostream>#include <assert.h>//解法1，递归void get_combine(std::vector<std::vector<std::string>>& src, std::vector<std::string>& result, std::vector<int> & index){assert(index.size() == src.size());std::string aline;for (int i=0; i<index.size(); ++i){aline += src[i][index[i]];if (i != index.size()-1){aline += "-";}}result.push_back(aline);//索引进行转动移位++index[index.size()-1];//const int lastColSize = src[index.size()-1].size();for (int i=index.size()-1; i>=0; --i){if (0 == index[i] % src[i].size()){if (i == 0){return; //达到上限，无法移动}index[i] = 0;++index[i-1]; //进位}else{break; //继续处理}}get_combine(src, result, index);}int _tmain(int argc, _TCHAR* argv[]){std::vector<std::string> v1, v2;v1.push_back("aa");v1.push_back("bb");v1.push_back("cc");v1.push_back("dd");v2.push_back("ee");v2.push_back("gg");v2.push_back("ff");v2.push_back("hh");std::vector<std::vector<std::string>> v;v.push_back(v1);v.push_back(v2);//std::vector<std::vector...中存的索引//类似密码表0001 0002，初始化为0000,然后//最低位达到该组的最大则上一位进一std::vector<int> index(v.size(), 0);std::vector<std::string> result;get_combine(v, result, index);//---组合结果输出-----for (int i=0; i<result.size(); ++i){std::cout<<result[i]<<std::endl;}//解决方案2，while循环std::vector<std::string> result2;for (int i=0; i<index.size(); ++i){index[i]  = 0;}bool bExit = false;//const int lastColSize = v[v.size()-1].size();while (true){std::string aline;for (int i=0; i<index.size(); ++i){aline += v[i][index[i]];if (i != index.size()-1){aline += "-";}}result2.push_back(aline);++index[index.size()-1];for (int i=index.size()-1; i>=0; --i){if (0 == index[i] % v[i].size()){if (i == 0){bExit = true;break; //达到上限，无法移动}index[i] = 0;++index[i-1]; //进位}else{break; //继续处理}}if (bExit){break;}}getchar();return 0;}