AGC002D Stamp Rally
うーん，解けそうな解法はできてたけど，定数倍が重かった
問題概要(原文)
$N$ 頂点 $M$ 本の辺からなる無向グラフが与えられる．
$Q$ 個のクエリが与えられる． $i$ 番目のクエリでは頂点 $u_i$ か $v_i$ を含む連結成分のサイズが $c$ 以上となるときの，そのような連結成分に含まれる辺の番号の最大値の最小値を求めよ．
考察
クエリが $1$ つの場合を考える．基本的にはクラスカル法のような感じで頂点同士をマージしていく． ここで，辺のコストが辺の番号なので，これは昇順に見るだけでよい． クエリが $1$ つの場合に解を二分探索すると $O(MlogM)$ で解ける．
クエリは $Q$ 個あったのであった． 二分探索の $k$ 回目のステップで， $i$ 番目のクエリは $(l_i + r_i)/2$ 番目の辺でいけるかどうかを調べる．
$1$ 回のステップでは $M$ 本の辺と， $Q$ 個のクエリについて調べる．よって計算量は $O((N+M+Q)logM)$ となる．
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 m = sc.Integer();
var a = new int[m];
var b = new int[m];
for (int i = 0; i < m; i++)
{
a[i] = sc.Integer() - 1;
b[i] = sc.Integer() - 1;
}
var q = sc.Integer();
var f = new int[q];
var t = new int[q];
var c = new int[q];
var l = new int[q];
var r = new int[q];
for (int i = 0; i < q; i++)
{
f[i] = sc.Integer() - 1;
t[i] = sc.Integer() - 1;
c[i] = sc.Integer();
l[i] = 0;
r[i] = m;
}
var A = Enumerate(m, x => new List<int>());
for (int k = 0; k < 18; k++)
{
var set = new DisjointSet(n);
for (int i = 0; i < q; i++)
{
var lr = (l[i] + r[i]) / 2;
A[lr].Add(i);
}
Debug.WriteLine(l.AsJoinedString());
Debug.WriteLine(r.AsJoinedString());
for (int i = 0; i < m; i++)
{
set.Unite(a[i], b[i]);
while (A[i].Count > 0)
{
var x = A[i][A[i].Count - 1];
A[i].RemoveAt(A[i].Count - 1);
int cnt = 0;
if (set.IsUnited(f[x], t[x]))
cnt = set.Size(f[x]);
else cnt = set.Size(f[x]) + set.Size(t[x]);
if (cnt >= c[x]) r[x] = i;
else l[x] = i;
}
}
}
for (int i = 0; i < q; i++)
IO.Printer.Out.WriteLine(r[i] + 1);
}
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
#region DisjointSet
public class DisjointSet
{
public int[] par;
byte[] ranks;
public DisjointSet(int n)
{
par = new int[n];
for (int i = 0; i < n; i++)
{
par[i] = -1;
}
ranks = new byte[n];
}
public int this[int id]
{
get
{
if ((par[id] < 0)) return id;
return par[id] = this[par[id]];
}
}
public bool Unite(int x, int y)
{
x = this[x]; y = this[y];
if (x == y) return false;
if (ranks[x] < ranks[y])
{
par[y] += par[x];
par[x] = y;
}
else
{
par[x] += par[y];
par[y] = x;
if (ranks[x] == ranks[y])
ranks[x]++;
}
return true;
}
public int Size(int x) { return -par[this[x]]; }
public bool IsUnited(int x, int y) { return this[x] == this[y]; }
}
#endregion
コメント
- クエリをまとめて二分探索というアイデアがなかった
- ロールバック平方分割はクッソ定数倍が重い
