ABC404-E Bowls and Beans

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("5");
            WillReturn.Add("1 1 2 1");
            WillReturn.Add("1 0 0 1");
            //3
        }
        else if (InputPattern == "Input2") {
            WillReturn.Add("6");
            WillReturn.Add("1 2 1 3 1");
            WillReturn.Add("1 1 0 1 1");
            //4
        }
        else if (InputPattern == "Input3") {
            WillReturn.Add("16");
            WillReturn.Add("1 1 1 2 5 1 1 3 4 1 4 3 1 1 2");
            WillReturn.Add("1 0 0 0 1 0 0 1 1 0 0 0 0 0 1");
            //7
        }
        else {
            string wkStr;
            while ((wkStr = Console.ReadLine()) != null) WillReturn.Add(wkStr);
        }
        return WillReturn;
    }

    static int[] mCArr;

    static void Main()
    {
        List<string> InputList = GetInputList();
        mCArr = InputList[1].Split(' ').Select(pX => int.Parse(pX)).ToArray();
        int[] AArr = InputList[2].Split(' ').Select(pX => int.Parse(pX)).ToArray();

        // 豆があるPosのList
        var BeanPosList = new List<int>();
        BeanPosList.Add(-1);

        for (int I = 0; I <= AArr.GetUpperBound(0); I++) {
            if (AArr[I] > 0) {
                BeanPosList.Add(I);
            }
        }

        int Answer = 0;
        int BeanPosList_UB = BeanPosList.Count - 1;
        for (int I = 0; I <= BeanPosList_UB - 1; I++) {
            int StaPos = BeanPosList[I + 1];
            int EndPos = BeanPosList[I];
            Answer += ExecDP(StaPos, EndPos);
        }
        Console.WriteLine(Answer);
    }

    // StaとEndを引数として、DPで移動コストを返す
    static int ExecDP(int pStaPos, int pEndPos)
    {
        // 最小コスト[現在位置]なDP表
        var PrevDP = new Dictionary<int, int>();
        PrevDP[pStaPos] = 0;

        int Answer = int.MaxValue;

        while (PrevDP.Count > 0) {
            var CurrDP = new Dictionary<int, int>();
            foreach (var EachPair in PrevDP) {
                int CurrPos = EachPair.Key;
                int MoveCnt = mCArr[CurrPos];

                for (int I = 1; I <= MoveCnt; I++) {
                    int ToPos = CurrPos - I;
                    if (ToPos < pEndPos) break;

                    int NewVal = EachPair.Value + 1;
                    if (NewVal >= Answer) {
                        continue;
                    }

                    // 枝切り
                    if (ToPos == pEndPos) {
                        Answer = Math.Min(Answer, NewVal);
                        continue;
                    }

                    if (CurrDP.ContainsKey(ToPos)) {
                        if (CurrDP[ToPos] <= NewVal) {
                            continue;
                        }
                    }
                    CurrDP[ToPos] = NewVal;
                }
            }
            PrevDP = CurrDP;
        }
        return Answer;
    }
}

解説

考察すると、
豆同士の間隔ごとに、DPで最小の移動コストを求めれば良いと分かります。