hdu6351 Beautiful Now 杭电第五场 暴力枚举

Beautiful Now

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

枚举每种交换情况，分别取最大和去最小，记得特判前导零

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e2+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
ll n, m;
bool cmp( char p, char q ) {
    return p > q;
}
string strmin, strmax, s, tmin, tmax;
void dfs1( ll x, ll cnt, string t ) {
    if( cnt > n-1 || x == t.length()-1 ) {
        //debug(cnt), debug(tmin), debug(strmin), debug(t);
        strmin = min(strmin,t);
        return ;
    }
    if( t[x] == tmin[x] ) {
        dfs1(x+1,cnt,t);
        return ;
    }
    char c = 'a';
    for( ll i = x+1; i < t.length(); i ++ ) {
        if( t[i] <= c ) {
            if( x == 0 && t[i] == '0' ) {
                continue;
            }
            c = t[i];
        }
    }
    if( c == 'a' ) {
        dfs1(x+1,cnt,t);
        return ;
    }
    for( ll i = x+1; i < t.length(); i ++ ) {
        if( t[i] == c ) {
            swap(t[i],t[x]);
            dfs1(x+1,cnt+1,t);
            swap(t[i],t[x]);
        }
    }
}
void dfs2( ll x, ll cnt, string t ) {
    if( cnt > n-1 || x == t.length()-1 ) {
        //debug(cnt), debug(tmax), debug(strmax), debug(t);
        strmax = max(strmax,t);
        return ;
    }
    if( t[x] == tmax[x] ) {
        dfs2(x+1,cnt,t);
        return ;
    }
    char c = '0';
    bool flag = true;
    for( ll i = x+1; i < t.length(); i ++ ) {
        if( t[i] >= c ) {
            c = t[i];
            flag = false;
        }
    }
    for( ll i = x+1; i < t.length(); i ++ ) {
        if( t[i] == c ) {
            swap(t[i],t[x]);
            dfs2(x+1,cnt+1,t);
            swap(t[i],t[x]);
        }
    }
}
string rev( string s ) {
    string t = "";
    for( ll i = 0, j = s.length()-1; i < s.length(); i ++, j -- ) {
        t = t + s[j];
    }
    return t;
}
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	ll T, t = 1;
	cin >> T;
	while( T -- ) {
        s = "";
        ll j = 0;
        cin >> m >> n;
        while(m) {
            char c = (m%10)+'0';
            s = s + c;
            m /= 10;
        }
        s = rev(s);
        tmin = s, tmax = s;
        sort(tmin.begin(),tmin.end());
        sort(tmax.begin(),tmax.end(),cmp);
        if( tmin[0] == '0' ) {
            char c = 'a';
            ll inx = -1;
            for( ll i = 1; i < tmin.length(); i ++ ) {
                if( tmin[i] != '0' && tmin[i] < c ) {
                    c = tmin[i];
                    inx = i;
                }
            }
            if( inx != -1 ) {
                swap(tmin[inx],tmin[0]);
            }
        }
        if( n >= s.length()-1 ) {
            cout << tmin << " " << tmax << endl;
        } else {
            strmin = s;
            dfs1(0,0,strmin);
            strmax = s;
            dfs2(0,0,strmax);
            cout << strmin << " " << strmax << endl;
        }
	}
	return 0;
}
/*
123112 2
111322 322111
10001 2
*/