• codeforce 124B——全排列dfs——Permutations


    You are given nk-digit integers. You have to rearrange the digits in the integers so that the difference between the largest and the smallest number was minimum. Digits should be rearranged by the same rule in all integers.

    Input

    The first line contains integers n and k — the number and digit capacity of numbers correspondingly (1 ≤ n, k ≤ 8). Next n lines containk-digit positive integers. Leading zeroes are 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 are rearranged in all integers by the same rule.

    Sample Input

    Input
    6 4
    5237
    2753
    7523
    5723
    5327
    2537
    Output
    2700
    Input
    3 3
    010
    909
    012
    Output
    3
    Input
    7 5
    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 will equal 5237 and 2537 correspondingly (they will 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 100, 99 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(int 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
    ",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
    ",anser);
        }
        return 0;
    }
    

      

  • 相关阅读:
    ch4-SPSS Statistics操作进阶
    ch3-SPSS Statistics操作入门
    概率论与数理统计-ch7-参数估计
    概率论与数理统计-ch6-样本与抽样分布
    概率论与数理统计-ch5-大数定律及中心极限定理
    概率论与数理统计-ch4-随机变量的数字特征
    概率论与数理统计-ch3-多维随机变量及其分布
    Maven的依赖冲突
    eclipse下的server插件
    JQuery了解
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4647009.html
Copyright © 2020-2023  润新知