// By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
//
// 3
// 7 4
// 2 4 6
// 8 5 9 3
//
// That is, 3 + 7 + 4 + 9 = 23.
//
// Find the maximum total from top to bottom of the triangle below:
//
// 75
// 95 64
// 17 47 82
// 18 35 87 10
// 20 04 82 47 65
// 19 01 23 75 03 34
// 88 02 77 73 07 63 67
// 99 65 04 28 06 16 70 92
// 41 41 26 56 83 40 80 70 33
// 41 48 72 33 47 32 37 16 94 29
// 53 71 44 65 25 43 91 52 97 51 14
// 70 11 33 28 77 73 17 78 39 68 17 57
// 91 71 52 38 17 14 91 43 58 50 27 29 48
// 63 66 04 68 89 53 67 30 73 16 69 87 40 31
// 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
//
// NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
using System;
using System.Collections.Generic;
using System.Text;
namespace projecteuler018
{
class Program
{
static void Main(string[] args)
{
F2();
}
#region F1()
private static void F1()
{
Console.WriteLine(new System.Diagnostics.StackTrace().GetFrame(0).GetMethod());
DateTime timeStart = DateTime.Now;
int[] data = new int[] { 75, 95, 64, 17, 47, 82, 18, 35, 87, 10, 20, 04, 82, 47, 65, 19, 01, 23, 75, 03, 34, 88, 02, 77, 73, 07, 63, 67, 99, 65, 04, 28, 06, 16, 70, 92, 41, 41, 26, 56, 83, 40, 80, 70, 33, 41, 48, 72, 33, 47, 32, 37, 16, 94, 29, 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, 04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23 };
Console.WriteLine(largestRoute1(data));
Console.WriteLine("Total Milliseconds is " + DateTime.Now.Subtract(timeStart).TotalMilliseconds + "\n\n");
}
private static int largestRoute1(int[] data)
{
if (data.Length == 1)
{
return data[0];
}
//获取层数
int layer = 1;
for (int i = data.Length; ; layer++)
{
i -= layer;
if (i == 0)
{
break;
}
}
//获取左下三角
//012345 -> 134
int[] leftData = new int[data.Length - layer];
int skipCount = 0;
int currentCount = 0;
int currentPosition = 0;
for (int i = 0; i < data.Length; i++)
{
if (currentCount-- > 0)
{
leftData[currentPosition++] = data[i];
}
else
{
skipCount++;
currentCount = skipCount;
}
}
//获取右下三角
//012345 -> 245
int[] rightData = new int[data.Length - layer];
skipCount = -1;
currentCount = 0;
currentPosition = 0;
for (int i = 0; i < data.Length; i++)
{
if (currentCount-- > 0)
{
rightData[currentPosition++] = data[i];
}
else
{
skipCount++;
currentCount = skipCount;
}
}
return Math.Max(largestRoute1(leftData), largestRoute1(rightData)) + data[0];
}
#endregion F1()
#region F2()
/// <summary>
/// 每个索引节点的最大路径值
/// </summary>
static int[] route;
/// <summary>
/// 数据总共的层数
/// </summary>
static int totalLayer;
private static void F2()
{
Console.WriteLine(new System.Diagnostics.StackTrace().GetFrame(0).GetMethod());
DateTime timeStart = DateTime.Now;
int[] data = new int[] { 75, 95, 64, 17, 47, 82, 18, 35, 87, 10, 20, 04, 82, 47, 65, 19, 01, 23, 75, 03, 34, 88, 02, 77, 73, 07, 63, 67, 99, 65, 04, 28, 06, 16, 70, 92, 41, 41, 26, 56, 83, 40, 80, 70, 33, 41, 48, 72, 33, 47, 32, 37, 16, 94, 29, 53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14, 70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57, 91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48, 63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31, 04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23 };
route = new int[data.Length];
Console.WriteLine(largestRoute2(data, 0, 0));
Console.WriteLine("Total Milliseconds is " + DateTime.Now.Subtract(timeStart).TotalMilliseconds + "\n\n");
}
/// <summary>
/// 获取某个三角的最大路径和
/// </summary>
/// <param name="data">三角数据</param>
/// <param name="layer">层数，从1开始</param>
/// <param name="index">在数组中的索引，从0开始，左下三角索引为index+currentlayer，右下三角的索引是index+currentlayer+1</param>
/// <returns></returns>
private static int largestRoute2(int[] data, int layer, int index)
{
if (layer == 1)
{
route[index] = data[0];
return data[0];
}
if (route[index] != 0 && layer != 0)
{
return route[index];
}
//如果是0层，则获取层数
if (layer == 0)
{
layer = 1;
for (int i = data.Length; ; layer++)
{
i -= layer;
if (i == 0)
{
break;
}
}
totalLayer = layer;
}
int currentLayer = totalLayer - layer + 1;
//获取左下三角
//012345 -> 134
int[] leftData = new int[data.Length - layer];
int skipCount = 0;
int currentCount = 0;
int currentPosition = 0;
for (int i = 0; i < data.Length; i++)
{
if (currentCount-- > 0)
{
leftData[currentPosition++] = data[i];
}
else
{
skipCount++;
currentCount = skipCount;
}
}
//获取右下三角
//012345 -> 245
int[] rightData = new int[data.Length - layer];
skipCount = -1;
currentCount = 0;
currentPosition = 0;
for (int i = 0; i < data.Length; i++)
{
if (currentCount-- > 0)
{
rightData[currentPosition++] = data[i];
}
else
{
skipCount++;
currentCount = skipCount;
}
}
route[index] = Math.Max(largestRoute2(leftData, layer - 1, index + currentLayer), largestRoute2(rightData, layer - 1, index + currentLayer + 1)) + data[0];
return route[index];
}
#endregion F2()
}
}
/*
Void F1()
1074
Total Milliseconds is 61.0077
Void F2()
1074
Total Milliseconds is 7.0009
By GodMoon
*/
