hdu6351---Beautiful Now（暴力）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/lpeaceminusone/article/details/81867589

Beautiful Now Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 2815    Accepted Submission(s): 1026   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

超时了无数次，竟然是因为标记数组应该用布尔型，CTRYB！

#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
#define maxn 100010
#define mod 7
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
char s[20];
int k,len;
int a[20];
bool vis[20];
bool check(){
    memset(vis,false,sizeof(vis));
    int tmp=0;
    for(int i=0;i<len;i++){
        if(vis[i])continue;
        int x=0;
        while(!vis[i]){
            x++;vis[i]=true;
            i=a[i];
        }
        tmp+=(x-1);
        if(tmp>k)return false;
    }
    return tmp<=k;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
    scanf("%s%d",s,&k);
    len=strlen(s);
    ll minn=inf,maxx=-1;
    int x=0;
    for(int i=0;i<len;i++){
            a[i]=i;
    }
    do{
        if(s[a[0]]!='0'&&check()){
             x=0;
            for(int i=0;i<len;i++)
                x=x*10+(s[a[i]]-'0');
        minn=minn>x?x:minn;
        maxx=maxx>x?maxx:x;
        }
        }while(next_permutation(a,a+len));
        printf("%lld %lld\n",minn,maxx);
    }
}