zoukankan      html  css  js  c++  java
  • 2.1.3 Sorting a Three-Valued Sequence

    Sorting a Three-Valued Sequence
    IOI'96 - Day 2

    Sorting is one of the most frequently performed computational tasks. Consider the special sorting problem in which the records to be sorted have at most three different key values. This happens for instance when we sort medalists of a competition according to medal value, that is, gold medalists come first, followed by silver, and bronze medalists come last.

    In this task the possible key values are the integers 1, 2 and 3. The required sorting order is non-decreasing. However, sorting has to be accomplished by a sequence of exchange operations. An exchange operation, defined by two position numbers p and q, exchanges the elements in positions p and q.

    You are given a sequence of key values. Write a program that computes the minimal number of exchange operations that are necessary to make the sequence sorted.

    PROGRAM NAME: sort3

    INPUT FORMAT

    Line 1: N (1 <= N <= 1000), the number of records to be sorted
    Lines 2-N+1: A single integer from the set {1, 2, 3}

    SAMPLE INPUT (file sort3.in)

    9
    2
    2
    1
    3
    3
    3
    2
    3
    1
    

    OUTPUT FORMAT

    A single line containing the number of exchanges required

    SAMPLE OUTPUT (file sort3.out)

    4
    
    {
    ID: makeeca1
    PROG: sort3
    LANG: PASCAL
    }
    var n,i,j,ans:longint;
        num:array[0..3]of longint;
        a,b:array[0..1000]of longint;
        cnt:array[0..3,0..3]of longint;
    begin
      assign(input,'sort3.in');reset(input);
      assign(output,'sort3.out');rewrite(output);
      readln(n);
      for i:=1 to n do
      begin
        readln(a[i]);
        inc(num[a[i]]);
      end;
      n:=0;
      for i:=1 to 3 do
        for j:=1 to num[i]do
        begin
          inc(n);
          b[n]:=i;
        end;
      fillchar(num,sizeof(num),0);
      for i:=1 to n do
      if b[i]<>a[i]then inc(cnt[b[i],a[i]]);
      if cnt[2,1]>cnt[1,2]then ans:=cnt[3,2]else ans:=cnt[2,3];
      inc(ans,cnt[2,1]+cnt[3,1]);
      writeln(ans);
      close(input); close(output);
    end.
  • 相关阅读:
    git 提交解决冲突(转载)
    impala系列: 时间函数
    impala系列: 字符串函数
    Impala系列: Impala常用的功能函数
    impala系列:impala特有的操作符
    impala系列: 同步Hive元数据和收集统计信息
    ETL脚本的版本管理方法和 SourceTree 使用
    几本不错的数据仓库和Hadoop书籍
    Kudu系列-基础
    sql parser
  • 原文地址:https://www.cnblogs.com/makeecat/p/3274550.html
Copyright © 2011-2022 走看看