Beautiful Now 【全排列+暴力枚举】

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

题目大意：给出一个数n，可以交换数中两个数字的位置，问交换k次以内得到的最下的数和最大的数

思路：用pos数组记录数组下标，运用全排列函数，得到每种下标的可能，然后判断；

具体看代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N=10;

int pos[N],k,n;
int a[N],cut;
int mark[N];

int judge()
{
    memset(mark,0,sizeof(mark));//记住每次要清0
    int tmp=0;
    for(int i=0;i<cut;i++)
    {
        if(mark[i])//如果这个下标被标记就跳过下面继续循环
            continue;
        int e=0;//交换的次数
        while(!mark[i])
        {
            e++;
            mark[i]=1;
            i=pos[i];//pos[i]为排列后的位置，i为排列前的位置
        }
        e--;//两两交换会多交换一次
        tmp+=e;
        if(tmp>k)
            return 0;
    }
    return 1;
}

int main()
{
    int  T;
    scanf("%d",&T);
    while(T--)
    {
        int i,j,temp,maxx=0,minn=inf;
        scanf("%d %d",&n,&k);
        cut=0;
        temp=n;
        while(temp>0)
        {
            a[cut++]=temp%10;
            temp/=10;
        }
        reverse(a,a+cut);
        for(i=0;i<cut;i++)//记录下标
            pos[i]=i;
        do
        {
            if(a[pos[0]]!=0&&judge())//注意前缀不能是0
            {
                int ans=0;
                for(i=0;i<cut;i++)
                    ans=ans*10+a[pos[i]];
                maxx=max(ans,maxx);
                minn=min(ans,minn);
            }
        }while(next_permutation(pos,pos+cut));//全排列函数，每次枚举每中下标的排列情况
        printf("%d %d\n",minn,maxx);
    }
    return 0;
}