HDU6351 Beautiful Now(2018HDU多校联赛第五场,思路)

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 ( x 1 x 2 ⋯ x m ) 10 satisfying that 1≤x1≤9 1 ≤ x 1 ≤ 9 , which means n=∑mi=1xi10m−i n = ∑ i = 1 m x i 10 m − i . In each swap, Anton can select two digits xi and xj (1≤i≤j≤m) ( 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 ≤ T ≤ 100 , 1≤n,k≤109 1 ≤ n , k ≤ 10 9

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

思路

给出t组数据，然后每组数据给出一个n和k，你每次可以交换任意两个数，总的交换次数不超过k次，问你在规定的次数之内这个数可以变成的最大值和最小值是多少(注意特判前导0).

因为一个数字顶多有9位，所以我们可以全排列暴力枚举排列数，复杂度就是 9! 9 ! ,每次枚举一个排列，计算出交换的次数如果符合条件就进行最大最小值处理。那么如何计算一个数交换几次可以变成另一个数，交换的数字必定会连成一个环，那么每个环需要交换的次数就是环内的元素个数减去1，(具体可以百度关键字使序列有序的最小交换次数)。

但是在本题中由于存在数字重复的问题，我们可以对于1到9所有的位数，提前预处理，存储一下经过多少次变换每个位置会变成什么，因为位置是不可能重复的。最后枚举一下从1~k的次数，然后转换成具体的数字求最大最小值。

代码

#include 
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
const int N = 1e6 + 10;
const int inf = 999999999;
vector<int> pre[10][10]; //pre[i][j]表示i位,交换次数为j的集合
int a[10], vis[10];
void init()
{
    for (int k = 1; k <= 9; k++)
    {
        for (int i = 1; i <= k; i++)
            a[i] = i;
        do
        {
            mem(vis, 0);
            int now = 0;
            for (int i = 1; i <= k; i++)
                now = now * 10 + a[i];
            int tmp = 0; //计算出变换需要的次数
            for (int i = 1; i <= k; i++)
                if (!vis[i])
                {
                    vis[i] = 1;
                    for (int j = a[i]; j != i; j = a[j])
                    {
                        vis[j] = 1;
                        tmp++;
                    }
                }
            pre[k][tmp].push_back(now);
        } while (next_permutation(a + 1, a + k + 1));
    }
}
int get_num(int n)
{
    int ans = 0;
    int tmp = 1;
    while (n)
    {
        ans += a[n % 10] * tmp;
        n /= 10;
        tmp *= 10;
    }
    return ans;
}
char s[20];
void solve()
{
    int k;
    scanf("%s%d", s, &k);
    int n = strlen(s);
    if (n == 10)
    {
        printf("1000000000 1000000000\n");
        return;
    }
    if (k >= n)
        k = n - 1;
    for (int i = 1; i <= n; i++)
        a[i] = s[i - 1] - '0';
    int maxx = 0, minn = inf;
    int l = 1;
    for (int i = 1; i < n; i++) //防止前导0
        l *= 10;
    for (int i = 0; i <= k; i++) //枚举所有的交换次数
    {
        for (auto num : pre[n][i])
        {
            int now = get_num(num); //根据位来获取值
            if (now < l)
                continue; //去除前导0
            maxx = max(maxx, now);
            minn = min(minn, now);
        }
    }
    printf("%d %d\n", minn, maxx);
}
int main()
{
    // freopen("in.txt", "r", stdin);
    init();
    int t;
    scanf("%d", &t);
    while (t--)
        solve();
    return 0;
}