[递归 搜索] Beautiful Now HDU6351

Beautiful Now

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1930    Accepted Submission(s): 730

 

Problem Description

Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits.
Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) and then swap them if the integer after this swap has no leading zero.
Could you please tell him the minimum integer and the maximum integer he can obtain after k swaps?

 

Input

The first line contains one integer T, indicating the number of test cases.
Each of the following T lines describes a test case and contains two space-separated integers n and k.
1≤T≤100, 1≤n,k≤109.

 

Output

For each test case, print in one line the minimum integer and the maximum integer which are separated by one space.

 

Sample Input

5

12 1

213 2

998244353 1

998244353 2

998244353 3

 

Sample Output

12 21

123 321

298944353 998544323

238944359 998544332

233944859 998544332

 

Source

2018 Multi-University Training Contest 5

#include <bits/stdc++.h>
using namespace std;

int a[20], b[20];
int len;
int Min, Max;
void dfs1(int s, int t, int k)
{
	if (k == 0)
	{
		int sum = 0;
		for (int i = 1; i <= len; i++)
			sum = sum * 10 + a[i];
		Min = min(Min, sum);
		return;
	}
	
	for (int i = s; i <= len; i++)
	{
		int t = a[i];
		for (int j = i; j <= len; j++)
		{
			if (i == 1 && a[j] == 0)
				continue;
			if (t > a[j])
				t = a[j];
		}  /// 从当前位起往后最小的值
		for (int j = i; j <= len; j++)
		{
			if (a[j] == t)
			{
				swap(a[i], a[j]); /// 交换一个满足条件的j
				dfs1(i + 1, len, k - 1);  /// 从i + 1开始到len交换k - 1次
				swap(a[i], a[j]); /// 还原 交换下一个满足条件的j
			}
		}
	}
}

void dfs2(int s, int t, int k)
{
	if (k == 0)
	{
		int sum = 0;
		for (int i = 1; i <= len; i++)
			sum = sum * 10 + b[i];   
		Max = max(Max, sum);
		return;
	}
	
	for (int i = s; i <= len; i++)
	{
		int t = b[i];
		for (int j = i; j <= len; j++)
		{
			if (t < b[j])
				t = b[j];
		}
		for (int j = i; j <= len; j++)
		{
			if (b[j] == t)
			{
				swap(b[i], b[j]);
				dfs2(i + 1, len, k - 1);
				swap(b[i], b[j]);
			}
		}
	}
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("D:\\in.txt", "r", stdin);
	#endif // ONLINE_JUDGE
	
	int T;
	scanf("%d", &T);
	while (T--)
	{
		char ch[20]; int k;
		scanf("%s %d", ch + 1, &k);
		len = strlen(ch + 1);
		
		for (int i = 1; i <= len; i++)
		{
			a[i] = ch[i] - '0';
			b[i] = a[i];
		}
		
		/// 数组长度为len, 最多交换len - 1次可得最值
		/// 如 len = 9, 234567891 交换8次得最小值123456789
		if (k >= len - 1)
		{
			sort(a + 1, a + len + 1);
			sort(b + 1, b + len + 1);
			
			int t = 1; 
			while (a[t] == 0)
				t++;
			swap(a[t], a[1]);
			for (int i = 1; i <= len; i++)
				printf("%d", a[i]);
			printf(" ");
			
			for (int i = len; i >= 1; i--)
				printf("%d", b[i]);
			printf("\n");
			
			continue;
		}
		
		Min = 0x7fffffff, Max = 0;
		dfs1(1, len, k);  /// 从1到len交换k次
		dfs2(1, len, k);
		printf("%d %d\n", Min, Max);
	}
	return 0;
}