【HDU - 6351】Beautiful Now 【暴力DFS + 剪枝】

本文链接：https://blog.youkuaiyun.com/qq_37383726/article/details/81479346

本文介绍了一种通过有限次数字位交换找到一个整数的最大和最小可能值的方法。利用DFS深度优先搜索策略，并通过适当的剪枝减少搜索空间，确保算法效率。

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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 $n = \sum_{i = 1}^{m}{x_i 10^{m - i}}$ In each swap, Anton can select two digits xi and xj $(1 \leq i \leq j \leq m)$ 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
题意：给一个数字x，问最多交换k次的任意两个位数的情况下，最大和最小的x（不能够有前导零)
分析：发现贪心贪不动，只能考虑dfs爆搜，很容易想到 $n!$ 的暴力做法，但是会T，这里要进行剪枝，如果是找最大，那么肯定比它大的才有交换的意义，最小同理。
代码

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

typedef long long ll;
typedef unsigned long long ull;

const int N = (int)1e5 + 11;
const int M = (int)15e6 + 11;
const int MOD = (int)1e9 + 7;
const int INF = (int) 0x3f3f3f3f;

string mx, mn;
void find_mx(string s, int st, int k){
    if(mx < s) mx = s;
    if(k == 0 || st == s.size()) return ;
    char ma = s[st];
    for(int i = st + 1; i < s.size(); i++) ma = max(ma,  s[i]);
    if(ma == s[st]) find_mx(s, st + 1, k);
    else {
        for(int i = st + 1; i < s.size(); i++) {
            if(ma == s[i]){
                swap(s[i], s[st]);
                find_mx(s, st + 1, k - 1);
                swap(s[i], s[st]);
            }
        }
    }
}

void find_mn(string s, int st, int k){
    if(mn > s) mn = s;
    if(k == 0 || st == s.size()) return ;
    char mi = s[st];
    for(int i = st + 1; i < s.size(); i++) {
        if(st == 0 && s[i] == '0') continue; // 前导零去掉
        mi = min(mi,  s[i]);
    }
    if(mi == s[st]) find_mn(s, st + 1, k);
    else {
        for(int i = st + 1; i < s.size(); i++) {
            if(mi == s[i]){
                swap(s[i], s[st]);
                find_mn(s, st + 1, k - 1);
                swap(s[i], s[st]);
            }
        }
    }
}

int main(){
    int _;for(scanf("%d", &_); _; _--){
        string s; int k; cin>>s >>k;
        mx = "0000000000", mn = "9999999999";
        find_mx(s, 0, k);
        find_mn(s, 0, k);
        cout<<mn<<" "<<mx<<"\n";
    }
return 0;
}