回溯法解组合问题——硬币问题

来源：互联网发布：centos 加载驱动编辑：程序博客网时间：2024/06/05 05:59

问题：

1，你有（足够的）5分，2分，1分的硬币，现在要凑出来12分的结果，那么最少的硬币组合是？

2，如果有5,12,7，-5，-7，-12这六个硬币。现在要你凑0-2000以内的任意数又该如何组合呢？

用回溯法解组合问题。在解空间中搜索。

// MoneyCount.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <vector>#include <iostream>#include <algorithm>#include <time.h>using namespace std;vector<int> MoneyKind;int Total=0;vector<int> bestSolution; //最优解int flag=0;//用来判断是否第一次找到可行解//solution 存有问题的当前解//total 目前还是需要多少//x 为增加哪一个硬币void FindSolution( vector<int> & solution, int total, int x ){if( total == x ) //找到了一个可行解{if( flag == 0 ){bestSolution = solution;flag = 1;}else{if( solution.size() < bestSolution.size() )bestSolution = solution;}return;}//搜索每一个子节点total = total - x;for( int i=0; i<(int)MoneyKind.size(); i++ ){solution.push_back( MoneyKind[i] );if( MoneyKind[i]<=total )//可行性约束{if( flag==1 && solution.size() <= bestSolution.size() )FindSolution( solution, total ,MoneyKind[i] ); if( flag==0)                                //没有找到可行解FindSolution( solution, total ,MoneyKind[i] );}solution.pop_back();}}void input(){int numKinds=0;cout << "How many kinds of coins?\n";cin >> numKinds;//MoneyKind.resize( numKinds );cout << "Input all kinds of coins:\n";for( int i=0; i<numKinds; i++){int m=0;cin >> m;MoneyKind.push_back(m);}//sort(MoneyKind.begin(), MoneyKind.end());cout << "Input the total money of coins:\n";cin >> Total;}int _tmain(int argc, _TCHAR* argv[]){input();clock_t begin = clock();vector<int> solution;for( int i=0; i<(int)MoneyKind.size(); i++ ){solution.push_back( MoneyKind[i] );if( MoneyKind[i]<=Total )//可行性约束{if( flag==1 && solution.size() <= bestSolution.size() )FindSolution( solution, Total ,MoneyKind[i] ); if( flag==0 )                                //没有找到可行解FindSolution( solution, Total ,MoneyKind[i] );}solution.pop_back();}if( flag == 0 )cout << "NO SOLUTION!" <<endl;else{cout << "SOLUTION IS:\n" ;sort( bestSolution.begin(),bestSolution.end() );for( int i=0; i<(int)bestSolution.size(); i++)cout << bestSolution[i] << " ";cout <<"\nTime used "<< (clock()-begin)/1000.0 << "s" <<endl;}return 0;}