1 金矿问题（Gold Mine Problem）
给定一个N*M尺寸的金矿，每个点都有一个非负数表示当前点所含的黄金数目，最开始矿工位于第一列，但是可以位于任意行。矿工只能向右，右上，右下三个方向移动。问该如何安排路线使得所挖的黄金的数量最多？
2 源程序（2种算法）：
using System;
using System.IO;
using System.Text;
using System.Collections;
using System.Collections.Generic;
namespace Legalsoft.Truffer.Algorithm
{
public static partial class Algorithm_Gallery
{
private static int Collect_Gold(int[,] gold, int r, int c, int n, int m)
{
// Base condition.
if ((r < 0) || (r == n) || (c == m))
{
return 0;
}
int rightUpperDiagonal = Collect_Gold(gold, r - 1, c + 1, n, m);
int right = Collect_Gold(gold, r, c + 1, n, m);
int rightLowerDiagonal = Collect_Gold(gold, r + 1, c + 1, n, m);
return gold[r, 0] + Math.Max(Math.Max(rightUpperDiagonal, rightLowerDiagonal), right);
}
public static int Get_Maxium_Gold(int[,] gold)
{
int n = gold.GetLength(0);
int m = gold.GetLength(1);
int maxGold = 0;
for (int i = 0; i < n; i++)
{
int goldCollected = Collect_Gold(gold, i, 0, n, m);
maxGold = Math.Max(maxGold, goldCollected);
}
return maxGold;
}
private static int Collect_Gold(int[,] gold, int r, int c, int n, int m, int[,] dp)
{
if ((r < 0) || (r == n) || (c == m))
{
return 0;
}
if (dp[r, 0] != -1)
{
return dp[r, 0];
}
int rightUpperDiagonal = Collect_Gold(gold, r - 1, c + 1, n, m, dp);
int right = Collect_Gold(gold, r, c + 1, n, m, dp);
int rightLowerDiagonal = Collect_Gold(gold, r + 1, c + 1, n, m, dp);
return gold[r, 0] + Math.Max(Math.Max(rightUpperDiagonal, rightLowerDiagonal), right);
}
public static int Get_Maxium_Gold_Second(int[,] gold)
{
int n = gold.GetLength(0);
int m = gold.GetLength(1);
int maxGold = 0;
int[,] dp = new int[n, m];
for (int row = 0; row < n; row++)
{
for (int col = 0; col < m; col++)
{
dp[row, col] = -1;
}
}
for (int i = 0; i < n; i++)
{
int goldCollected = Collect_Gold(gold, i, 0, n, m, dp);
maxGold = Math.Max(maxGold, goldCollected);
}
return maxGold;
}
}
}
3 源程序
using System;
using System.IO;
using System.Text;
using System.Collections;
using System.Collections.Generic;
namespace Legalsoft.Truffer.Algorithm
{
public static partial class Algorithm_Gallery
{
private static int Collect_Gold(int[,] gold, int r, int c, int n, int m)
{
// Base condition.
if ((r < 0) || (r == n) || (c == m))
{
return 0;
}
int rightUpperDiagonal = Collect_Gold(gold, r - 1, c + 1, n, m);
int right = Collect_Gold(gold, r, c + 1, n, m);
int rightLowerDiagonal = Collect_Gold(gold, r + 1, c + 1, n, m);
return gold[r, 0] + Math.Max(Math.Max(rightUpperDiagonal, rightLowerDiagonal), right);
}
public static int Get_Maxium_Gold(int[,] gold)
{
int n = gold.GetLength(0);
int m = gold.GetLength(1);
int maxGold = 0;
for (int i = 0; i < n; i++)
{
int goldCollected = Collect_Gold(gold, i, 0, n, m);
maxGold = Math.Max(maxGold, goldCollected);
}
return maxGold;
}
private static int Collect_Gold(int[,] gold, int r, int c, int n, int m, int[,] dp)
{
if ((r < 0) || (r == n) || (c == m))
{
return 0;
}
if (dp[r, 0] != -1)
{
return dp[r, 0];
}
int rightUpperDiagonal = Collect_Gold(gold, r - 1, c + 1, n, m, dp);
int right = Collect_Gold(gold, r, c + 1, n, m, dp);
int rightLowerDiagonal = Collect_Gold(gold, r + 1, c + 1, n, m, dp);
return gold[r, 0] + Math.Max(Math.Max(rightUpperDiagonal, rightLowerDiagonal), right);
}
public static int Get_Maxium_Gold_Second(int[,] gold)
{
int n = gold.GetLength(0);
int m = gold.GetLength(1);
int maxGold = 0;
int[,] dp = new int[n, m];
for (int row = 0; row < n; row++)
{
for (int col = 0; col < m; col++)
{
dp[row, col] = -1;
}
}
for (int i = 0; i < n; i++)
{
int goldCollected = Collect_Gold(gold, i, 0, n, m, dp);
maxGold = Math.Max(maxGold, goldCollected);
}
return maxGold;
}
}
}
