AGC003C BBuBBBlesort!
これすき
問題概要(原文)
$N$ 要素の pairwise distinct な数列 $A$ がある．
隣接要素の交換のコストが $1$ ，距離 $2$ の要素の交換がコスト $0$ でできる．
昇順に並び替えるコストの最小値を求めよ．
考察
$A$ を昇順に並び替えた数列 $B$ を考える．
$A_{i} = B_j$ のとき， $A_i = j \; \text{mod} \; 2$ としてもよい．
これはどういうことかというと，距離 $2$ のスワップは何度でもできるので，そのようなものを区別しなくてもよい．
その後 $A_{i} \neq i \; \text{mod} \; 2$ である数を数えて $2$ で割った値が答え． これは要素 $N$ の数列が $2$ つあって， $A_{i}$ と $B_{i}$ をコスト $1$ で交換するか，$A_{i}$ と $A_{i+1}$ をコスト $0$ で交換するかのどちらかができる，とすると分かりやすい．
ソースコード
using System;
using System.Linq;
using System.Linq.Expressions;
using System.Collections.Generic;
using Debug = System.Diagnostics.Debug;
using StringBuilder = System.Text.StringBuilder;
using System.Numerics;
using Number = System.Int64;
namespace Program
{
public class Solver
{
public void Solve()
{
var n = sc.Integer();
var a = sc.Integer(n);
var b = a.ToList(); b.Sort();
for (int i = 0; i < n; i++)
a[i] = b.BinarySearch(a[i]) % 2;
Debug.WriteLine(a.AsJoinedString());
var ans = 0;
for (int i = 0; i < n; i++)
if (i % 2 == 0 && a[i] == 1) ans++;
IO.Printer.Out.WriteLine(ans);
}
public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());
static T[] Enumerate<T>(int n, Func<int, T> f) { var a = new T[n]; for (int i = 0; i < n; ++i) a[i] = f(i); return a; }
static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
}
}
#region main
static class Ex
{
static public string AsString(this IEnumerable<char> ie) { return new string(System.Linq.Enumerable.ToArray(ie)); }
static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") { return string.Join(st, ie); }
static public void Main()
{
var solver = new Program.Solver();
solver.Solve();
Program.IO.Printer.Out.Flush();
}
}
#endregion
#region Ex
namespace Program.IO
{
using System.IO;
using System.Text;
using System.Globalization;
public class Printer: StreamWriter
{
static Printer() { Out = new Printer(Console.OpenStandardOutput()) { AutoFlush = false }; }
public static Printer Out { get; set; }
public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
public Printer(System.IO.Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
public Printer(System.IO.Stream stream, Encoding encoding) : base(stream, encoding) { }
public void Write<T>(string format, T[] source) { base.Write(format, source.OfType<object>().ToArray()); }
public void WriteLine<T>(string format, T[] source) { base.WriteLine(format, source.OfType<object>().ToArray()); }
}
public class StreamScanner
{
public StreamScanner(Stream stream) { str = stream; }
public readonly Stream str;
private readonly byte[] buf = new byte[1024];
private int len, ptr;
public bool isEof = false;
public bool IsEndOfStream { get { return isEof; } }
private byte read()
{
if (isEof) return 0;
if (ptr >= len) { ptr = 0; if ((len = str.Read(buf, 0, 1024)) <= 0) { isEof = true; return 0; } }
return buf[ptr++];
}
public char Char() { byte b = 0; do b = read(); while ((b < 33 || 126 < b) && !isEof); return (char)b; }
public string Scan()
{
var sb = new StringBuilder();
for (var b = Char(); b >= 33 && b <= 126; b = (char)read())
sb.Append(b);
return sb.ToString();
}
public string ScanLine()
{
var sb = new StringBuilder();
for (var b = Char(); b != '\n'; b = (char)read())
if (b == 0) break;
else if (b != '\r') sb.Append(b);
return sb.ToString();
}
public long Long()
{
if (isEof) return long.MinValue;
long ret = 0; byte b = 0; var ng = false;
do b = read();
while (b != 0 && b != '-' && (b < '0' || '9' < b));
if (b == 0) return long.MinValue;
if (b == '-') { ng = true; b = read(); }
for (; true; b = read())
{
if (b < '0' || '9' < b)
return ng ? -ret : ret;
else ret = ret * 10 + b - '0';
}
}
public int Integer() { return (isEof) ? int.MinValue : (int)Long(); }
public double Double() { var s = Scan(); return s != "" ? double.Parse(s, CultureInfo.InvariantCulture) : double.NaN; }
private T[] enumerate<T>(int n, Func<T> f)
{
var a = new T[n];
for (int i = 0; i < n; ++i) a[i] = f();
return a;
}
public char[] Char(int n) { return enumerate(n, Char); }
public string[] Scan(int n) { return enumerate(n, Scan); }
public double[] Double(int n) { return enumerate(n, Double); }
public int[] Integer(int n) { return enumerate(n, Integer); }
public long[] Long(int n) { return enumerate(n, Long); }
}
}
#endregion
コメント
- こういうの思いつくのすごいなぁ
