【发布时间】：2017-09-19 12:08:30
【问题描述】：
我试图在具有连续数字的矩阵中找到给出正确答案的最长路径。函数调用递归执行，直到附近没有连续数字，并且每次检查单元格是否被访问
#include<bits/stdc++.h>
#define n 3
using namespace std;
// Returns length of the longest path beginning with mat[i][j].
// This function mainly uses lookup table dp[n][n]
int findLongestFromACell(int i, int j, int mat[n][n], int dp[n][n])
{
// Base case
if (i<0 || i>=n || j<0 || j>=n)
return 0;
// If this subproblem is already solved
if (dp[i][j] != -1)
return dp[i][j];
// Since all numbers are unique and in range from 1 to n*n,
// there is atmost one possible direction from any cell
if (j<n-1 && ((mat[i][j] +1) == mat[i][j+1]))
return dp[i][j] = 1 + findLongestFromACell(i,j+1,mat,dp);
if (j>0 && (mat[i][j] +1 == mat[i][j-1]))
return dp[i][j] = 1 + findLongestFromACell(i,j-1,mat,dp);
if (i>0 && (mat[i][j] +1 == mat[i-1][j]))
return dp[i][j] = 1 + findLongestFromACell(i-1,j,mat,dp);
if (i<n-1 && (mat[i][j] +1 == mat[i+1][j]))
return dp[i][j] = 1 + findLongestFromACell(i+1,j,mat,dp);
// If none of the adjacent fours is one greater
return dp[i][j] = 1;
}
// Returns length of the longest path beginning with any cell
int finLongestOverAll(int mat[n][n])
{
int result = 1; // Initialize result
// Create a lookup table and fill all entries in it as -1
int dp[n][n];
memset(dp, -1, sizeof dp);
// Compute longest path beginning from all cells
for (int i=0; i<n; i++)
{
for (int j=0; j<n; j++)
{
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
// Update result if needed
result = max(result, dp[i][j]);
}
}
return result;
}
// Driver program
int main()
{
int mat[n][n] = {{1, 10, 9},
{5, 3, 8},
{4, 6, 7}};
cout << "Length of the longest path is "
<< finLongestOverAll(mat);
return 0;
}
但是当我尝试相同的代码来查找二进制矩阵中的最长路径时，程序停止执行
#include<bits/stdc++.h>
#define n 3
using namespace std;
// Returns length of the longest path beginning with mat[i][j].
// This function mainly uses lookup table dp[n][n]
int findLongestFromACell(int i, int j, int mat[n][n], int dp[n][n])
{
// Base case
if (i<0 || i>=n || j<0 || j>=n)
return 0;
// If this subproblem is already solved
if (dp[i][j] != -1)
return dp[i][j];
// Since all numbers are unique and in range from 1 to n*n,
// there is atmost one possible direction from any cell
if (j<n-1 && (1 == mat[i][j+1]))
return dp[i][j] = 1 + findLongestFromACell(i,j+1,mat,dp);
if (j>0 && (1 == mat[i][j-1]))
return dp[i][j] = 1 + findLongestFromACell(i,j-1,mat,dp);
if (i>0 && (1 == mat[i-1][j]))
return dp[i][j] = 1 + findLongestFromACell(i-1,j,mat,dp);
if (i<n-1 && (1 == mat[i+1][j]))
return dp[i][j] = 1 + findLongestFromACell(i+1,j,mat,dp);
// If none of the adjacent fours is one greater
return dp[i][j] = 1;
}
// Returns length of the longest path beginning with any cell
int finLongestOverAll(int mat[n][n])
{
int result = 1; // Initialize result
// Create a lookup table and fill all entries in it as -1
int dp[n][n];
memset(dp, -1, sizeof dp);
// Compute longest path beginning from all cells
for (int i=0; i<n; i++)
{
for (int j=0; j<n; j++)
{
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
// Update result if needed
result = max(result, dp[i][j]);
}
}
return result;
}
// Driver program
int main()
{
int mat[n][n] = {{1, 0, 0},
{1, 0, 0},
{1, 1, 1}};
cout << "Length of the longest path is "
<< finLongestOverAll(mat);
return 0;
}
此代码中的错误是什么。提前致谢
【问题讨论】：
-
请edit您的问题提供minimal reproducible example。
-
不包含
，它是一个私有的非标准标头，不包含在内。不要#define 整数常量。 -
您确定它停止工作了吗？你是怎么确定的？也许执行需要很长时间才能完成。
-
@FrançoisAndrieux 只需要几个 epoch 就可以无限递归最终溢出堆栈？
-
这段代码看起来很糟糕。如果您将这个几乎 C 重构为使用 C++ 标准库容器，您可以使用
.at()方法（而不是operator[]）并在不使用调试器的情况下捕获潜在的索引错误。即使没有，重构代码也会对其他读者有所帮助。
标签： c++ recursion matrix longest-path