Beautiful Now（dfs）（思维）

                                      Beautiful Now

                       Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
T                                    otal Submission(s): 1719    Accepted Submission(s): 646

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

Recommend

chendu   |   We have carefully selected several similar problems for you:  6361 6360 6359 6358 6357 

题意：每次交换两个数字（可以自己交换自己），最多交换k次，问交换后的最大最小值是多少。

每次找出当前序列中的所有的最小值，与最大数位交换，如果最大数位是最小值则开始下一位，最大值同理。

多校的时候一开始忽略了自己交换自己的情况，写的很复杂，后来修改了之后又遗漏了交换后要枚举最小值所在的不同位置，思路是对的，很可惜最后没有A掉这道题。

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

char n[20],ans[20],s[20];
int k,T,len,tp;

void dfs(int now,int lef)
{
    if(now>=len||lef==0)
        {if(strcmp(ans,s)>0) strcpy(ans,s);return;}
    int temp=now;
    for(int i=now;i<=len;i++)
        if(s[temp]>s[i]) temp=i;
    if(temp==now)
        dfs(now+1,lef);
    else
    {
        for(int i=now;i<=len;i++)
        {
            if(s[i]==s[temp])
            {
                swap(s[i],s[now]);
                dfs(now+1,lef-1);
                swap(s[i],s[now]);
            }
        }
    }
}
void dfsmax(int now,int lef)
{
    if(now>=len||lef==0)
        {if(strcmp(ans,s)<0) strcpy(ans,s);return;}
    int temp=now;
    for(int i=now;i<=len;i++)
        if(s[temp]<s[i]) temp=i;
    if(temp==now)
        dfsmax(now+1,lef);
    else
    {
        for(int i=now;i<=len;i++)
        {
            if(s[i]==s[temp])
            {
                swap(s[i],s[now]);
                dfsmax(now+1,lef-1);
                swap(s[i],s[now]);
            }
        }

    }
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%[^ ] %d",n,&k);
        len=strlen(n+1);
        strcpy(ans,n);
        strcpy(s,n);
        {//最小值
            tp=1;
            for(int i=1;i<=len;i++)
                if(n[i]<n[tp]&&n[i]!='0')
                    tp=i;
            if(tp==1)
                dfs(2,k);
            else
            {
                for(int i=1;i<=len;i++)
                    if(s[i]==s[tp])
                    {
                        swap(s[i],s[1]);
                        dfs(2,k-1);
                        swap(s[i],s[1]);
                    }
            }
            printf("%s ",ans+1);
        }
        {//最大值
            strcpy(ans,n);
            dfsmax(1,k);
            printf("%s\n",ans+1);
        }
    }
    return 0;
}