ABC282-E Choose Two and Eat One

問題へのリンク

C#のソース

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static string InputPattern = "InputX";

    static List<string> GetInputList()
    {
        var WillReturn = new List<string>();

        if (InputPattern == "Input1") {
            WillReturn.Add("4 10");
            WillReturn.Add("4 2 3 2");
            //20
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("20 100");
            WillReturn.Add("29 31 68 20 83 66 23 84 69 96 41 61 83 37 52 71 18 55 40 8");
            //1733
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static long mN;
    static long mM;
    static long[] mAArr;

    struct EdgeInfoDef
    {
        internal long FromNode;
        internal long ToNode;
        internal long Cost;
    }
    static List<EdgeInfoDef> mEdgeInfoList = new List<EdgeInfoDef>();

    static void Main()
    {
        List<string> InputList = GetInputList();
        long[] wkArr = InputList[0].Split(' ').Select(pX => long.Parse(pX)).ToArray();
        mN = wkArr[0];
        mM = wkArr[1];

        mAArr = InputList[1].Split(' ').Select(pX => long.Parse(pX)).ToArray();

        for (long I = 0; I <= mAArr.GetUpperBound(0); I++) {
            for (long J = I + 1; J <= mAArr.GetUpperBound(0); J++) {
                EdgeInfoDef WillAdd;
                WillAdd.FromNode = I + 1;
                WillAdd.ToNode = J + 1;

                long Kou1 = DeriveBekijyou(mAArr[I], mAArr[J], mM);
                long Kou2 = DeriveBekijyou(mAArr[J], mAArr[I], mM);
                WillAdd.Cost = (Kou1 + Kou2) % mM;
                mEdgeInfoList.Add(WillAdd);
            }
        }

        // クラスカル法で解く
        var InsUnionFind = new UnionFind();
        for (long I = 1; I <= mN; I++) {
            InsUnionFind.MakeSet(I);
        }

        // 枝をコストの降順にソート
        mEdgeInfoList.Sort((a, b) => b.Cost.CompareTo(a.Cost));

        long SumCost = 0;
        foreach (EdgeInfoDef EachEdgeInfo in mEdgeInfoList) {
            long FromNode = EachEdgeInfo.FromNode;
            long ToNode = EachEdgeInfo.ToNode;
            long RootNode1 = InsUnionFind.FindSet(EachEdgeInfo.FromNode);
            long RootNode2 = InsUnionFind.FindSet(EachEdgeInfo.ToNode);
            if (RootNode1 != RootNode2) {
                InsUnionFind.Unite(FromNode, ToNode);
                SumCost += EachEdgeInfo.Cost;
            }
        }
        Console.WriteLine(SumCost);
    }

    // 繰り返し2乗法で、(NのP乗) Mod Mを求める
    static long DeriveBekijyou(long pN, long pP, long pM)
    {
        long CurrJyousuu = pN % pM;
        long CurrShisuu = 1;
        long WillReturn = 1;

        while (true) {
            // 対象ビットが立っている場合
            if ((pP & CurrShisuu) > 0) {
                WillReturn = (WillReturn * CurrJyousuu) % pM;
            }

            CurrShisuu *= 2;
            if (CurrShisuu > pP) return WillReturn;
            CurrJyousuu = (CurrJyousuu * CurrJyousuu) % pM;
        }
    }
}

#region UnionFind
// UnionFindクラス
internal class UnionFind
{
    private class NodeInfoDef
    {
        internal long ParentNode;
        internal long Rank;
    }
    private Dictionary<long, NodeInfoDef> mNodeInfoDict =
        new Dictionary<long, NodeInfoDef>();

    // 要素が1つである木を森に追加
    internal void MakeSet(long pNode)
    {
        NodeInfoDef WillAdd = new NodeInfoDef();
        WillAdd.ParentNode = pNode;
        WillAdd.Rank = 0;
        mNodeInfoDict[pNode] = WillAdd;
    }

    // 合併処理
    internal void Unite(long pX, long pY)
    {
        long XNode = FindSet(pX);
        long YNode = FindSet(pY);
        long XRank = mNodeInfoDict[XNode].Rank;
        long YRank = mNodeInfoDict[YNode].Rank;

        if (XRank > YRank) {
            mNodeInfoDict[YNode].ParentNode = XNode;
        }
        else {
            mNodeInfoDict[XNode].ParentNode = YNode;
            if (XRank == YRank) {
                mNodeInfoDict[YNode].Rank++;
            }
        }
    }

    // ノードを引数として、木の根を取得
    internal long FindSet(long pTargetNode)
    {
        // 根までの経路上のノードのList
        var PathNodeList = new List<long>();

        long CurrNode = pTargetNode;
        while (CurrNode != mNodeInfoDict[CurrNode].ParentNode) {
            PathNodeList.Add(CurrNode);
            CurrNode = mNodeInfoDict[CurrNode].ParentNode;
        }

        // 経路圧縮 (親ポインタの付け替え)
        foreach (long EachPathNode in PathNodeList) {
            mNodeInfoDict[EachPathNode].ParentNode = CurrNode;
        }
        return CurrNode;
    }

    internal void DebugPrint()
    {
        foreach (var EachPair in mNodeInfoDict.OrderBy(pX => pX.Key)) {
            Console.WriteLine("mNodeInfoDict[{0}].ParentNode={1}",
                EachPair.Key, EachPair.Value.ParentNode);
        }
    }
}
#endregion

解説

考察すると、
全域木の中でも枝のコストの総合計が最大の木が解だと分かります。