回溯法解组合问题——硬币问题

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;
}