2018 Multi-University Training Contest 5 B （搜索)

Beautiful Now

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 4513    Accepted Submission(s): 457

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

 

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题目大意：给出一个不超过10e9的数，要求交换k次后求出最小值和最大值。两个问题相互独立，自己可以和自己交换。

解题思路：我们尝试了一下贪心，发现对于许多情况都无法解决，索性来了一个暴力的搜索。就是如果当前位后边有比它小的

我们就交换这两个数字，然后用这个数字重新去搜。找最大的和最小的方法一样。最后就是交换次数的控制，因为我感觉这个如果我们每次都交换两个数的话，在不超过数字长度的情况下一定能搜到最小的，所以就取搜的次数最小的那个。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
using namespace std;
#define LL long long
int cnt=1;
int mm=0x3f3f3f3f;
int ma=-1;
int ans[15];
int ans1[15];

void dfs(int k)
{
    if(k==0)return ;
    dfs(k/10);
    ans[cnt++]=k%10;
}

void dfs1(int k)
{
    if(k==0)return ;
    dfs1(k/10);
    ans1[cnt++]=k%10;
}

int cal()
{
    int sum=0;
    for(int i=1;i<cnt;i++)
    {
        sum=sum*10+ans[i];
    }
    return sum;
}

int cal1()
{
    int sum=0;
    for(int i=1;i<cnt;i++)
    {
        sum=sum*10+ans1[i];
    }
    return sum;
}

void dfsmin(int k,int step,int ci)
{
    //cout<<k<<" "<<step<<" "<<ci<<endl;
    mm=min(k,mm);
    if(step>min(ci,cnt))return ;
    for(int i=step+1;i<cnt;i++)
    {
       if(step==1&&ans[i]==0)continue;
       if(ans[i]<ans[step])
       {
           swap(ans[i],ans[step]);
           int tmp=cal();
           dfsmin(tmp,step+1,ci);
           swap(ans[i],ans[step]);

       }
    }
    dfsmin(k,step+1,ci+=1);
}
void dfsmax(int k,int step,int ci)
{
    ma=max(k,ma);
    if(step>min(ci,cnt))return ;
    for(int i=step+1;i<cnt;i++)
    {
       if(step==1&&ans1[i]==0)continue;
       if(ans1[i]>ans1[step])
       {
           swap(ans1[i],ans1[step]);
           int tmp=cal1();
           dfsmax(tmp,step+1,ci);
           swap(ans1[i],ans1[step]);

       }
    }
    dfsmax(k,step+1,ci+=1);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,ci;
        cin>>n>>ci;
        dfs(n);
        cnt=1;
        dfs1(n);
        dfsmin(n,1,ci);
        cout<<mm;
        dfsmax(n,1,ci);
        cout<<" "<<ma<<endl;
        cnt=1;
        mm=0x3f3f3f3f;
        ma=-1;
    }
}