给定一个大小为N x N的矩阵grid[][] ，任务是找到从左上角到达矩阵右下角所需的最小成本，其中移动到新单元格的成本是[S /2] + K ，其中S是前一个索引处的分数， K是当前索引处的矩阵元素。
注意：这里， [X] 是不超过X的最大整数。
例子：
Input: grid[][] = {{0, 3, 9, 6}, {1, 4, 4, 5}, {8, 2, 5, 4}, {1, 8, 5, 9}}
Output: 12
Explanation: One of the possible set of moves is as follows 0 -> 1 -> 4 -> 4 -> 7 -> 7 -> 12.
Input: grid[][] = {{0, 82, 2, 6, 7}, {4, 3, 1, 5, 21}, {6, 4, 20, 2, 8}, {6, 6, 64, 1, 8}, {1, 65, 1, 6, 4}}
Output: 7
处理方法：按照以下步骤解决问题：
- 使矩阵的第一个元素为零。
- 在[0, N-1]范围内遍历：
- 初始化一个列表，比如moveList ，并附加移动i和j 。
- 初始化另一个列表，比如可能的方式，并在其中附加moveList 。
- 初始化一个列表，比如可能的WaysSum，最初是一个空列表。
- 遍历列表可能的方式：
- 遍历附加的moveList ：
- 检查移动是否等于i，然后更新i = i + 1。
- 否则，更新j = j + 1 。
- 初始化一个变量，比如tempSum。设置tempSum = tempSum + grid[i][j] 。
- 在循环后的可能方式列表中附加tempSum 。
- 遍历附加的moveList ：
- 打印来自possibleWaysSum的最小成本作为输出。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
vector > possibleWays;
// Returns true if str[curr] does not matches with any
// of the characters after str[start]
bool shouldSwap(char str[], int start, int curr)
{
for (int i = start; i < curr; i++) {
if (str[i] == str[curr]) {
return false;
}
}
return true;
}
// Function for the swap
void swap(char str[], int i, int j)
{
char c = str[i];
str[i] = str[j];
str[j] = c;
}
// Prints all distinct permutations in str[0..n-1]
void findPermutations(char str[], int index, int n)
{
if (index >= n) {
vector l;
for (int i = 0; i < n; i++)
l.push_back(str[i]);
possibleWays.push_back(l);
return;
}
for (int i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the characters
// after str[index]
bool check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function to print the minimum cost
void minCost(int grid[][5], int N)
{
vector moveList;
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
vector possibleWaysSum;
for (int k = 0; k < N - 1; k++) {
moveList.push_back('i');
moveList.push_back('j');
possibleWays.clear();
// Convert into set to make only unique values
int n = moveList.size();
char str[n];
for (int i = 0; i < n; i++)
str[i] = moveList[i];
// To store the unique permutation of moveLst
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum.clear();
// Traverse the list
for (vector way : possibleWays) {
int i = 0, j = 0, tempSum = 0;
for (char move : way) {
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum = (int)(floor(tempSum / 2))
+ grid[i][j];
}
possibleWaysSum.push_back(tempSum);
}
}
// Print the minimum possible cost
int ans = possibleWaysSum[0];
for (int i = 1; i < possibleWaysSum.size(); i++)
ans = min(ans, possibleWaysSum[i]);
cout << ans;
}
// Driven Program
int main()
{
// Size of the grid
int N = 4;
// Given grid[][]
int grid[][5] = { { 0, 3, 9, 6 },
{ 1, 4, 4, 5 },
{ 8, 2, 5, 4 },
{ 1, 8, 5, 9 } };
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
return 0;
}
// This code is contributed by Kingash. Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{
static ArrayList > possibleWays;
// Returns true if str[curr] does not matches with any
// of the characters after str[start]
static boolean shouldSwap(char str[], int start,
int curr)
{
for (int i = start; i < curr; i++)
{
if (str[i] == str[curr])
{
return false;
}
}
return true;
}
// Prints all distinct permutations in str[0..n-1]
static void findPermutations(char str[], int index,
int n)
{
if (index >= n)
{
ArrayList l = new ArrayList<>();
for (char ch : str)
l.add(ch);
possibleWays.add(l);
return;
}
for (int i = index; i < n; i++) {
// Proceed further for str[i] only if it
// doesn't match with any of the characters
// after str[index]
boolean check = shouldSwap(str, index, i);
if (check) {
swap(str, index, i);
findPermutations(str, index + 1, n);
swap(str, index, i);
}
}
}
// Function for the swap
static void swap(char[] str, int i, int j)
{
char c = str[i];
str[i] = str[j];
str[j] = c;
}
// Function to print the minimum cost
static void minCost(int grid[][], int N)
{
ArrayList moveList = new ArrayList<>();
// Making top-left value 0
// implicitly to generate list of moves
grid[0][0] = 0;
ArrayList possibleWaysSum
= new ArrayList<>();
for (int k = 0; k < N - 1; k++) {
moveList.add('i');
moveList.add('j');
possibleWays = new ArrayList<>();
// Convert into set to make only unique values
int n = moveList.size();
char str[] = new char[n];
for (int i = 0; i < n; i++)
str[i] = moveList.get(i);
// To store the unique permutation of moveLst
// into the possibleWays
findPermutations(str, 0, n);
possibleWaysSum = new ArrayList<>();
// Traverse the list
for (ArrayList way : possibleWays) {
int i = 0, j = 0, tempSum = 0;
for (char move : way) {
if (move == 'i') {
i += 1;
}
else {
j += 1;
}
// Stores cost according to given
// conditions
tempSum = (int)(Math.floor(tempSum / 2))
+ grid[i][j];
}
possibleWaysSum.add(tempSum);
}
}
// Print the minimum possible cost
int ans = possibleWaysSum.get(0);
for (int i = 1; i < possibleWaysSum.size(); i++)
ans = Math.min(ans, possibleWaysSum.get(i));
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
// Size of the grid
int N = 4;
// Given grid[][]
int grid[][] = { { 0, 3, 9, 6 },
{ 1, 4, 4, 5 },
{ 8, 2, 5, 4 },
{ 1, 8, 5, 9 } };
// Function call to print the minimum
// cost to reach bottom-right corner
// from the top-left corner of the matrix
minCost(grid, N);
}
}
// This code is contributed by Kingash. Python3
# Python3 program for the above approach
from itertools import permutations
from math import floor
# Function to print the minimum cost
def minCost(grid, N):
moveList = []
# Making top-left value 0
# implicitly to generate list of moves
grid[0][0] = 0
for i in range(N - 1):
moveList.append('i')
moveList.append('j')
# Convert into set to make only unique values
possibleWays = list(set(permutations(moveList)))
possibleWaysSum = []
# Traverse the list
for way in possibleWays:
i, j, tempSum = 0, 0, 0
for move in way:
if move == 'i':
i += 1
else:
j += 1
# Stores cost according to given conditions
tempSum = floor(tempSum/2) + grid[i][j]
possibleWaysSum.append(tempSum)
minWayIndex = possibleWaysSum.index(min(possibleWaysSum))
# Print the minimum possible ccost
print(min(possibleWaysSum))
# Size of the grid
N = 4
# Given grid[][]
grid = [[0, 3, 9, 6], [1, 4, 4, 5], [8, 2, 5, 4], [1, 8, 5, 9]]
# Function call to print the minimum
# cost to reach bottom-right corner
# from the top-left corner of the matrix
minCost(grid, N)输出：
12时间复杂度： O(N 2 )
辅助空间： O(N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live