找零钱(回溯)

收银员现有 n 张面值分别为 v1,v2,…,vn 的纸币。若找零金额为 m，则一共有多少种找零方法？
注：0<n≤1000，0<v​1,v2,…,vn≤10000，0<m≤10000
输入格式
n v1,v2,…,vn m
输出格式
若有解，则输出全部找零方案，每输出一种 若无解，则输出“None”

输入样例1
6
3 1 4 3 2 7
9
输出样例1
3 1 3 2
3 4 2
4 3 2
2 7
输入样例2
5
5 3 4 6 7
2
输出样例2
None

解法一:

#include <iostream>
#include <algorithm>
using namespace std;
//典型回溯  dfs 
int m, n, n_num, sum = 0, v[1024], book[1024] = {0};//p记录状态 
 
void println()
{
	int flag = false;
	for(int i = 1; i <= n; ++i)
	{
		if(book[i]) 
		{
			if(flag) putchar(' ');
			printf("%d", v[i]);
			flag = true;	
		}	
	}
	putchar('\n');
	++sum;	
} 

void Solveways(int i)
{
	if(n_num < m)
	{
		while(i <= n)
		{			
			if(!book[i])
			{
				book[i] = 1;
				n_num += v[i];
				Solveways(i + 1);
				book[i] = 0;
				n_num -= v[i];				
			}
			++i;
		}
	}
	else if(n_num == m) println();	
} 

int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &v[i]);
	scanf("%d", &m);	
	Solveways(1);
	if(sum == 0) printf("None\n"); 
	return 0;	
} 

解法二(组合 + 剪枝):

#include <iostream>
#include <algorithm>
using namespace std;
int m, n, n_num, sum = 0, v[1024], x[1024] = {0};//p记录状态
int sums = 0, lefts = 0; 
// 满二叉树
// 剪枝函数就是为了避免不必要的重复运算 
int Prune(int i, int k)
{
	if(i == 0) if(sums + v[k] > m)         return 0;
	if(i == 1) if(sums + lefts - v[k] < m) return 0;
	return 1;
}
 
int Sum()
{
	int sum = 0;
	for(int i = 1; i <= n; ++i) if(x[i] == 0) sum += v[i];
	return sum;
}
 
void OutPut()
{
	int i, j; 
	for(i = 1; i <= n; ++i) 
	if(x[i] == 0)
	{
		printf("%d", v[i]); 
		break;	
	}
	for(j = i + 1; j <= n; ++j) if(x[j] == 0) printf(" %d", v[j]);
	putchar('\n');
	++sum;
} 

void Solveways(int k)
{
	if(k > n)
	{
		if(Sum() == m) OutPut();
		return ;
	}
	for(int i = 0; i <= 1; ++i)// 无剪枝
	if(Prune(i, k)) 
	{
		if(i == 0) sums += v[k];
		lefts -= v[k];
		x[k] = i;
		Solveways(k + 1);
		if(i == 0) sums -= v[k];
		lefts += v[k];		
	}
} 

int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) scanf("%d", &v[i]);
	scanf("%d", &m);
	for(int i = 1; i <= n; ++i) lefts += v[i];	
	Solveways(1);
	if(sum == 0) printf("None\n"); 
	return 0;	
}
/*
6
3 1 4 3 2 7
9
*/