Codeforce 124b--Full arrangement dfs--permutations

Source: Internet
Author: User

You are given nk-digit integers. Rearrange the digits in the integers so, the difference between the largest and the smallest number was MI Nimum. Digits should is rearranged by the same rule in all integers.

Input

The first line contains integers n and K -the number and digit capacity of numbers Correspondin Gly (1≤ n, k ≤8). Next n lines containk-digit positive integers. Leading zeroes is allowed both in the initial integers and the integers resulting from the rearranging of digits.

Output

Print a single number:the minimally possible difference between the largest and the smallest number after the digits is Rearranged in all integers by the same rule.

Sample Input

Input
6 4
5237
2753
7523
5723
5327
2537
Output
2700
Input
7 ·
010
909
012
Output
3
Input
9 {
50808
36603
37198
44911
29994
42543
50156
Output
20522

Hint

In the first sample, if we rearrange the digits in numbers as (3,1,4,2), then the 2-nd and the 4-th numbers would equal 523 7 and 2537 correspondingly (they would be maximum and minimum for such order of digits).

In the second sample, if we swap the second digits and the first ones, we get integers, and 102.

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int inf = 0x3f3f3f3f ; const int INF = -0x3f3f3f3f;char a[10][10];int b[10][10];int c[10];int vis[10];int ans[10];int N, k;int anser;void dfs (i       NT CNT) {if (cnt = = k+1) {memset (c,0,sizeof (c));               for (int i = 1, i <= N; i++) {for (int j = 1; j <= K; j + +) {C[i] = c[i]*10 + b[i][ans[j]];      }} sort (c + 1, C + n + 1);       printf ("%d%d\n", c[n],c[1]);      Anser = min (Anser, C[n]-c[1]);     return;            } for (int i = 1; I <= K; i++) {if (vis[i] = = 0) {ans[cnt] = i;            Vis[i] = 1;            DFS (CNT+1);        Vis[i] = 0;        }}}int Main () {while (~scanf ("%d%d", &n,&k)) {anser = inf;        for (int i = 1; I <= n; i++) scanf ("%s", &a[i]); for (int i = 1, i <= N; i++) {for (int j = 0;j < K; J + +) {b[i][j+1] = a[i][j]-' 0 ';           }}/* for (int i = 1, i <= N; i++) {for (int j = 1; j <= K; j + +)          printf ("%d", b[i][j]);        Puts ("");    }*/memset (vis,0,sizeof (VIS));    DFS (1);    printf ("%d\n", Anser); } return 0;}

  

Codeforce 124b--Full arrangement dfs--permutations

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.