找出从(0，0)到(N-1，M-1)
的最小差路径
原文:https://www . geeksforgeeks . org/find-最小差异路径-从 0-0 到 n-1-m-1/
给定两个 2D 阵列 b[][] 和c[][]N行和 M 列。任务是沿着从 (0，0) 到(N–1，M–1)的路径，最小化 b[i][j]s 之和与 c[i][j]s 之和的绝对差。 示例:
输入: b[][] = {{1，4}，{2，4}}，c[][] = {{3，2}，{3，1}} 输出: 0 选择路径(0，0) - > (1，0) - > (1，1) b[I][j]s 的和为= 1+2+4 = 7 c[I][j]s 的和为= 3 + 3 + 1 = 7 【T8
方法:答案独立于决定 b[i][j] 和 c[i][j] 的顺序和路径。因此，让我们考虑一个布尔表，如果 (i，j) 能够以 k 的最小差异达到，则 dp[i][j][k] 将为真。 如果是真的，那么对于单元格 (i + 1，j) 来说，要么是 k + |b i+1，j–cI+1，j | 要么是| k –| bI+1，j–cI+1，j|。正方形 (i，j + 1) 也是如此。因此，表格可以按照 i 和 j 的递增顺序填写。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAXI 50
int dp[MAXI][MAXI][MAXI * MAXI];
int n, m;
// Function to return the minimum difference
// path from (0, 0) to (N - 1, M - 1)
int minDifference(int x, int y, int k,
vector<vector<int> > b,
vector<vector<int> > c)
{
// Terminating case
if (x >= n or y >= m)
return INT_MAX;
// Base case
if (x == n - 1 and y == m - 1) {
int diff = b[x][y] - c[x][y];
return min(abs(k - diff), abs(k + diff));
}
int& ans = dp[x][y][k];
// If it is already visited
if (ans != -1)
return ans;
ans = INT_MAX;
int diff = b[x][y] - c[x][y];
// Recursive calls
ans = min(ans, minDifference(x + 1, y,
abs(k + diff), b, c));
ans = min(ans, minDifference(x, y + 1,
abs(k + diff), b, c));
ans = min(ans, minDifference(x + 1, y,
abs(k - diff), b, c));
ans = min(ans, minDifference(x, y + 1,
abs(k - diff), b, c));
// Return the value
return ans;
}
// Driver code
int main()
{
n = 2, m = 2;
vector<vector<int> > b = { { 1, 4 }, { 2, 4 } };
vector<vector<int> > c = { { 3, 2 }, { 3, 1 } };
memset(dp, -1, sizeof(dp));
// Function call
cout << minDifference(0, 0, 0, b, c);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
final static int MAXI = 50 ;
static int dp[][][] = new int[MAXI][MAXI][MAXI * MAXI];
static int n, m;
final static int INT_MAX = Integer.MAX_VALUE;
// Function to return the minimum difference
// path from (0, 0) to (N - 1, M - 1)
static int minDifference(int x, int y, int k,
int b[][], int c[][])
{
// Terminating case
if (x >= n || y >= m)
return INT_MAX;
// Base case
if (x == n - 1 && y == m - 1)
{
int diff = b[x][y] - c[x][y];
return Math.min(Math.abs(k - diff),
Math.abs(k + diff));
}
int ans = dp[x][y][k];
// If it is already visited
if (ans != -1)
return ans;
ans = INT_MAX;
int diff = b[x][y] - c[x][y];
// Recursive calls
ans = Math.min(ans, minDifference(x + 1, y,
Math.abs(k + diff), b, c));
ans = Math.min(ans, minDifference(x, y + 1,
Math.abs(k + diff), b, c));
ans = Math.min(ans, minDifference(x + 1, y,
Math.abs(k - diff), b, c));
ans = Math.min(ans, minDifference(x, y + 1,
Math.abs(k - diff), b, c));
// Return the value
return ans;
}
// Driver code
public static void main (String[] args)
{
n = 2; m = 2;
int b[][] = { { 1, 4 }, { 2, 4 } };
int c[][] = { { 3, 2 }, { 3, 1 } };
for(int i = 0; i < MAXI; i++)
{
for(int j = 0; j < MAXI; j++)
{
for(int k = 0; k < MAXI * MAXI; k++)
{
dp[i][j][k] = -1;
}
}
}
// Function call
System.out.println(minDifference(0, 0, 0, b, c));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
import numpy as np
import sys
MAXI = 50
INT_MAX = sys.maxsize
dp = np.ones((MAXI, MAXI, MAXI * MAXI));
dp *= -1
# Function to return the minimum difference
# path from (0, 0) to (N - 1, M - 1)
def minDifference(x, y, k, b, c) :
# Terminating case
if (x >= n or y >= m) :
return INT_MAX;
# Base case
if (x == n - 1 and y == m - 1) :
diff = b[x][y] - c[x][y];
return min(abs(k - diff), abs(k + diff));
ans = dp[x][y][k];
# If it is already visited
if (ans != -1) :
return ans;
ans = INT_MAX;
diff = b[x][y] - c[x][y];
# Recursive calls
ans = min(ans, minDifference(x + 1, y,
abs(k + diff), b, c));
ans = min(ans, minDifference(x, y + 1,
abs(k + diff), b, c));
ans = min(ans, minDifference(x + 1, y,
abs(k - diff), b, c));
ans = min(ans, minDifference(x, y + 1,
abs(k - diff), b, c));
# Return the value
return ans;
# Driver code
if __name__ == "__main__" :
n = 2; m = 2; b = [ [ 1, 4 ], [ 2, 4 ] ];
c = [ [ 3, 2 ], [ 3, 1 ] ];
# Function call
print(minDifference(0, 0, 0, b, c));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG
{
static int MAXI = 50 ;
static int [,,]dp = new int[MAXI, MAXI,
MAXI * MAXI];
static int n, m;
static int INT_MAX = int.MaxValue;
// Function to return the minimum difference
// path from (0, 0) to (N - 1, M - 1)
static int minDifference(int x, int y, int k,
int [,]b, int [,]c)
{
int diff = 0;
// Terminating case
if (x >= n || y >= m)
return INT_MAX;
// Base case
if (x == n - 1 && y == m - 1)
{
diff = b[x, y] - c[x, y];
return Math.Min(Math.Abs(k - diff),
Math.Abs(k + diff));
}
int ans = dp[x, y, k];
// If it is already visited
if (ans != -1)
return ans;
ans = INT_MAX;
diff = b[x, y] - c[x, y];
// Recursive calls
ans = Math.Min(ans, minDifference(x + 1, y,
Math.Abs(k + diff), b, c));
ans = Math.Min(ans, minDifference(x, y + 1,
Math.Abs(k + diff), b, c));
ans = Math.Min(ans, minDifference(x + 1, y,
Math.Abs(k - diff), b, c));
ans = Math.Min(ans, minDifference(x, y + 1,
Math.Abs(k - diff), b, c));
// Return the value
return ans;
}
// Driver code
public static void Main ()
{
n = 2; m = 2;
int [,]b = { { 1, 4 }, { 2, 4 } };
int [,]c = { { 3, 2 }, { 3, 1 } };
for(int i = 0; i < MAXI; i++)
{
for(int j = 0; j < MAXI; j++)
{
for(int k = 0; k < MAXI * MAXI; k++)
{
dp[i, j, k] = -1;
}
}
}
// Function call
Console.WriteLine(minDifference(0, 0, 0, b, c));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// JavaScript implementation of the approach
let MAXI = 50 ;
let dp = new Array(MAXI);
let n, m;
let INT_MAX = Number.MAX_VALUE;
// Function to return the minimum difference
// path from (0, 0) to (N - 1, M - 1)
function minDifference(x, y, k, b, c)
{
// Terminating case
if (x >= n || y >= m)
return INT_MAX;
// Base case
if (x == n - 1 && y == m - 1)
{
let diff = b[x][y] - c[x][y];
return Math.min(Math.abs(k - diff),
Math.abs(k + diff));
}
let ans = dp[x][y][k];
// If it is already visited
if (ans != -1)
return ans;
ans = INT_MAX;
let diff = b[x][y] - c[x][y];
// Recursive calls
ans = Math.min(ans, minDifference(x + 1, y,
Math.abs(k + diff), b, c));
ans = Math.min(ans, minDifference(x, y + 1,
Math.abs(k + diff), b, c));
ans = Math.min(ans, minDifference(x + 1, y,
Math.abs(k - diff), b, c));
ans = Math.min(ans, minDifference(x, y + 1,
Math.abs(k - diff), b, c));
// Return the value
return ans;
}
n = 2; m = 2;
let b = [ [ 1, 4 ], [ 2, 4 ] ];
let c = [ [ 3, 2 ], [ 3, 1 ] ];
for(let i = 0; i < MAXI; i++)
{
dp[i] = new Array(MAXI);
for(let j = 0; j < MAXI; j++)
{
dp[i][j] = new Array(MAXI * MAXI);
for(let k = 0; k < MAXI * MAXI; k++)
{
dp[i][j][k] = -1;
}
}
}
// Function call
document.write(minDifference(0, 0, 0, b, c));
</script>
Output:
0
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