给定一个数字网格，找到最大长度的Snake序列并打印出来。如果存在多个最大长度的蛇形序列，请打印其中的任何一个。
蛇形序列由网格中的相邻数字组成，因此对于每个数字，右侧的数字或其下方的数字为其值的+1或-1。例如，如果您位于网格中的位置(x，y)，则可以向右移动，即(x，y + 1)(如果该数字为±1)，也可以向下移动，即(x + 1，y)，如果该数字是±1。
例如，
9，6，5，2
8、7、6、5
7,3，1,6
1，1，1，7
在上面的网格中，最长的蛇序列为：(9、8、7、6、5、6、7)
下图显示了所有可能的路径–
强烈建议您最小化浏览器，然后自己尝试。
这个想法是使用动态编程。对于矩阵的每个单元，我们保留一条在当前单元中结束的蛇的最大长度。最大长度的蛇序列将具有最大值。最大值单元格将对应于蛇的尾巴。为了打印蛇，我们需要从尾巴一直回溯到蛇的头部。
令T [i] [i]代表一条蛇的最大长度，该蛇在单元格(i，j)处结束，那么对于给定的矩阵M，DP关系定义为：
T [0] [0] = 0
如果M [i] [j] = M [i] [j – 1]±1，则T [i] [j] = max(T [i] [j]，T [i] [j – 1] + 1)
如果M [i] [j] = M [i – 1] [j]±1，则T [i] [j] = max(T [i] [j]，T [i – 1] [j] + 1)
以下是该想法的实现–
C++
// C++ program to find maximum length
// Snake sequence and print it
#include
using namespace std;
#define M 4
#define N 4
struct Point
{
int x, y;
};
// Function to find maximum length Snake sequence path
// (i, j) corresponds to tail of the snake
list findPath(int grid[M][N], int mat[M][N],
int i, int j)
{
list path;
Point pt = {i, j};
path.push_front(pt);
while (grid[i][j] != 0)
{
if (i > 0 &&
grid[i][j] - 1 == grid[i - 1][j])
{
pt = {i - 1, j};
path.push_front(pt);
i--;
}
else if (j > 0 &&
grid[i][j] - 1 == grid[i][j - 1])
{
pt = {i, j - 1};
path.push_front(pt);
j--;
}
}
return path;
}
// Function to find maximum length Snake sequence
void findSnakeSequence(int mat[M][N])
{
// table to store results of subproblems
int lookup[M][N];
// initialize by 0
memset(lookup, 0, sizeof lookup);
// stores maximum length of Snake sequence
int max_len = 0;
// store cordinates to snake's tail
int max_row = 0;
int max_col = 0;
// fill the table in bottom-up fashion
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
{
// do except for (0, 0) cell
if (i || j)
{
// look above
if (i > 0 &&
abs(mat[i - 1][j] - mat[i][j]) == 1)
{
lookup[i][j] = max(lookup[i][j],
lookup[i - 1][j] + 1);
if (max_len < lookup[i][j])
{
max_len = lookup[i][j];
max_row = i, max_col = j;
}
}
// look left
if (j > 0 &&
abs(mat[i][j - 1] - mat[i][j]) == 1)
{
lookup[i][j] = max(lookup[i][j],
lookup[i][j - 1] + 1);
if (max_len < lookup[i][j])
{
max_len = lookup[i][j];
max_row = i, max_col = j;
}
}
}
}
}
cout << "Maximum length of Snake sequence is: "
<< max_len << endl;
// find maximum length Snake sequence path
list path = findPath(lookup, mat, max_row,
max_col);
cout << "Snake sequence is:";
for (auto it = path.begin(); it != path.end(); it++)
cout << endl << mat[it->x][it->y] << " ("
<< it->x << ", " << it->y << ")" ;
}
// Driver code
int main()
{
int mat[M][N] =
{
{9, 6, 5, 2},
{8, 7, 6, 5},
{7, 3, 1, 6},
{1, 1, 1, 7},
};
findSnakeSequence(mat);
return 0;
} Java
// Java program to find maximum length
// Snake sequence and print it
import java.util.*;
class GFG
{
static int M = 4;
static int N = 4;
static class Point
{
int x, y;
public Point(int x, int y)
{
this.x = x;
this.y = y;
}
};
// Function to find maximum length Snake sequence path
// (i, j) corresponds to tail of the snake
static List findPath(int grid[][],
int mat[][],
int i, int j)
{
List path = new LinkedList<>();
Point pt = new Point(i, j);
path.add(0, pt);
while (grid[i][j] != 0)
{
if (i > 0 &&
grid[i][j] - 1 == grid[i - 1][j])
{
pt = new Point(i - 1, j);
path.add(0, pt);
i--;
}
else if (j > 0 && grid[i][j] - 1 ==
grid[i][j - 1])
{
pt = new Point(i, j - 1);
path.add(0, pt);
j--;
}
}
return path;
}
// Function to find maximum length Snake sequence
static void findSnakeSequence(int mat[][])
{
// table to store results of subproblems
int [][]lookup = new int[M][N];
// initialize by 0
// stores maximum length of Snake sequence
int max_len = 0;
// store cordinates to snake's tail
int max_row = 0;
int max_col = 0;
// fill the table in bottom-up fashion
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
{
// do except for (0, 0) cell
if (i != 0 || j != 0)
{
// look above
if (i > 0 &&
Math.abs(mat[i - 1][j] -
mat[i][j]) == 1)
{
lookup[i][j] = Math.max(lookup[i][j],
lookup[i - 1][j] + 1);
if (max_len < lookup[i][j])
{
max_len = lookup[i][j];
max_row = i; max_col = j;
}
}
// look left
if (j > 0 &&
Math.abs(mat[i][j - 1] -
mat[i][j]) == 1)
{
lookup[i][j] = Math.max(lookup[i][j],
lookup[i][j - 1] + 1);
if (max_len < lookup[i][j])
{
max_len = lookup[i][j];
max_row = i; max_col = j;
}
}
}
}
}
System.out.print("Maximum length of Snake " +
"sequence is: " + max_len + "\n");
// find maximum length Snake sequence path
List path = findPath(lookup, mat, max_row,
max_col);
System.out.print("Snake sequence is:");
for (Point it : path)
System.out.print("\n" + mat[it.x][it.y] + " (" +
it.x + ", " + it.y + ")");
}
// Driver code
public static void main(String[] args)
{
int mat[][] = {{9, 6, 5, 2},
{8, 7, 6, 5},
{7, 3, 1, 6},
{1, 1, 1, 7}};
findSnakeSequence(mat);
}
}
// This code is contributed by 29AjayKumar Python3
# Python program to find maximum length
# Snake sequence and print it
M = 4
N = 4
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
# Function to find maximum length Snake sequence path
# (i, j) corresponds to tail of the snake
def findPath(grid, mat, i, j):
path = list()
pt = Point(i, j)
path.append(pt)
while (grid[i][j] != 0):
if (i > 0 and grid[i][j]-1 == grid[i-1][j]):
pt = Point(i-1, j)
path.append(pt)
i -= 1
elif (j > 0 and grid[i][j]-1 == grid[i][j-1]):
pt = Point(i, j-1)
path.append(pt)
j -= 1
return path
# Function to find maximum length Snake sequence
def findSnakeSequence(mat):
# table to store results of subproblems
# initialize by 0
lookup = [[0 for i in range(N)] for j in range(M)]
# stores maximum length of Snake sequence
max_len = 0
# store cordinates to snake's tail
max_row = 0
max_col = 0
# fill the table in bottom-up fashion
for i in range(M):
for j in range(N):
# do except for (0, 0) cell
if (i or j):
# look above
if (i > 0 and
abs(mat[i-1][j] - mat[i][j]) == 1):
lookup[i][j] = max(lookup[i][j],
lookup[i-1][j] + 1)
if (max_len < lookup[i][j]):
max_len = lookup[i][j]
max_row = i
max_col = j
# look left
if (j > 0 and
abs(mat[i][j-1] - mat[i][j]) == 1):
lookup[i][j] = max(lookup[i][j],
lookup[i][j-1] + 1)
if (max_len < lookup[i][j]):
max_len = lookup[i][j]
max_row = i
max_col = j
print("Maximum length of Snake sequence is:", max_len)
# find maximum length Snake sequence path
path = findPath(lookup, mat, max_row, max_col)
print("Snake sequence is:")
for ele in reversed(path):
print(mat[ele.x][ele.y],
" (", ele.x, ", ", ele.y, ")", sep = "")
# Driver code
mat = [[9, 6, 5, 2],
[8, 7, 6, 5],
[7, 3, 1, 6],
[1, 1, 1, 7]]
findSnakeSequence(mat)
# This code is contributed
# by Soumen GhoshC#
// C# program to find maximum length
// Snake sequence and print it
using System;
using System.Collections.Generic;
class GFG
{
static int M = 4;
static int N = 4;
public class Point
{
public int x, y;
public Point(int x, int y)
{
this.x = x;
this.y = y;
}
};
// Function to find maximum length Snake sequence path
// (i, j) corresponds to tail of the snake
static List findPath(int [,]grid,
int [,]mat,
int i, int j)
{
List path = new List();
Point pt = new Point(i, j);
path.Insert(0, pt);
while (grid[i, j] != 0)
{
if (i > 0 &&
grid[i, j] - 1 == grid[i - 1, j])
{
pt = new Point(i - 1, j);
path.Insert(0, pt);
i--;
}
else if (j > 0 && grid[i, j] - 1 ==
grid[i, j - 1])
{
pt = new Point(i, j - 1);
path.Insert(0, pt);
j--;
}
}
return path;
}
// Function to find maximum length Snake sequence
static void findSnakeSequence(int [,]mat)
{
// table to store results of subproblems
int [,]lookup = new int[M, N];
// initialize by 0
// stores maximum length of Snake sequence
int max_len = 0;
// store cordinates to snake's tail
int max_row = 0;
int max_col = 0;
// fill the table in bottom-up fashion
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
{
// do except for (0, 0) cell
if (i != 0 || j != 0)
{
// look above
if (i > 0 &&
Math.Abs(mat[i - 1, j] -
mat[i, j]) == 1)
{
lookup[i, j] = Math.Max(lookup[i, j],
lookup[i - 1, j] + 1);
if (max_len < lookup[i,j])
{
max_len = lookup[i, j];
max_row = i; max_col = j;
}
}
// look left
if (j > 0 &&
Math.Abs(mat[i, j - 1] -
mat[i, j]) == 1)
{
lookup[i, j] = Math.Max(lookup[i, j],
lookup[i, j - 1] + 1);
if (max_len < lookup[i, j])
{
max_len = lookup[i, j];
max_row = i; max_col = j;
}
}
}
}
}
Console.Write("Maximum length of Snake " +
"sequence is: " + max_len + "\n");
// find maximum length Snake sequence path
List path = findPath(lookup, mat, max_row,
max_col);
Console.Write("Snake sequence is:");
foreach (Point it in path)
Console.Write("\n" + mat[it.x,it.y] +
" (" + it.x + ", " + it.y + ")");
}
// Driver code
public static void Main(String[] args)
{
int [,]mat = {{9, 6, 5, 2},
{8, 7, 6, 5},
{7, 3, 1, 6},
{1, 1, 1, 7}};
findSnakeSequence(mat);
}
}
// This code is contributed by Princi Singh 输出 ：
Maximum length of Snake sequence is: 6
Snake sequence is:
9 (0, 0)
8 (1, 0)
7 (1, 1)
6 (1, 2)
5 (1, 3)
6 (2, 3)
7 (3, 3)上述解决方案的时间复杂度为O(M * N)。上述解决方案使用的辅助空间为O(M * N)。如果不需要打印蛇，则可以将空间进一步减小为O(N)，因为我们仅使用最后一行的结果。
参考：堆栈溢出