HDU 6351 - Beautiful Now（DFS）

Beautiful Now

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1055    Accepted Submission(s): 364

 

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

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))

const int MAXN = 1e5+10;
const int INF = 0x3f3f3f3f;

int T,k,len;
string s,str;
int maxx ,minn;
int change(string s){
    int num = 0;
    for(int i=0;i<s.size();i++){
        num = num * 10 + (s[i]-'0');
    }
    return num;
}
void findmaxx(string s, int id, int k) {
    maxx = max(maxx,change(s));
    if(k == 0 || id == len) return ;
    char t = s[id];
    for(int j = id + 1; j < len; ++j)
        t = max(t, s[j]);
    if(t == s[id])
        findmaxx(s, id + 1, k);
    else {
        for(int j = id + 1; j < len; ++j) {
            if(s[j] == t) {
                swap(s[id], s[j]);
                findmaxx(s, id + 1, k - 1);
                swap(s[id], s[j]);
            }
        }
    }
}
void findminn(string s, int id, int k) {
    minn = min(minn, change(s));
    if(k == 0 || id == len) return ;
    char t = s[id];
    for(int j = id + 1; j < len; ++j)
        t = min(t, (id == 0 && s[j] == '0' ? '9' : s[j]));
    if(t == s[id])
        findminn(s, id + 1, k);
    else {
        for(int j = id + 1; j < len; ++j) {
            if(s[j] == t) {
                swap(s[id], s[j]);
                findminn(s, id + 1, k - 1);
                swap(s[id], s[j]);
            }
        }
    }
}
int main(){
    scanf("%d",&T);
    while(T--){
        cin>>s>>k;
        maxx = 0;minn = INF;
        len = s.size();
        str = s;
        findmaxx(s,0,k);
        findminn(s,0,k);
        printf("%d %d\n",minn,maxx);
    }
    return 0;
}