给定一个由字符组成的NXN矩阵。还给出了 Q 查询，其中每个查询包含一个坐标 (X, Y)。对于每个查询，通过垂直或水平移动找到从 (1, 1) 到 (X, Y) 的所有路径，并取具有最大数量的路径在里面。任务是沿该路径打印非“a”字符的数量。
例子：
Input: mat[][] = {{'a', 'b', 'a'},
{'a', 'c', 'd'},
{'b', 'a', 'b'}}
Queries:
X = 1, Y = 3
X = 3, Y = 3
Output:
1st query: 1
2nd query: 2
Query-1: There is only one path from (1, 1) to (1, 3)
i.e., "aba" and the number of characters
which are not 'a' is 1.
Query-2: The path which has the maximum number of 'a'
in it is "aabab", hence non 'a'
characters are 2. 该问题是 Min-Cost 路径 Dp 问题的变体。我们需要预先计算 DP[][] 数组，然后答案将是 DP[X][Y]，因此每个查询都可以在 O(1) 中得到回答。如果索引位置 (1, 1)字符不是 ‘a’，则将 dp[1][1] 值增加 1。然后简单地迭代行和列，并考虑 ‘a’ 做一个最小成本 DP 1 和非 ‘a’字符为 0。由于 DP[][] 数组存储从 (1, 1) 到任何索引 (i, j) 的最小成本路径，因此可以在 O(1) 中回答查询)。
下面是上述方法的实现：
C++
// C++ program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
#include
using namespace std;
const int n = 3;
int dp[n][n];
// Function to answer queries
void answerQueries(pair queries[], int q)
{
// Iterate till query
for (int i = 0; i < q; i++) {
// Decrease to get 0-based indexing
int x = queries[i].first;
x--;
int y = queries[i].second;
y--;
// Print answer
cout << dp[x][y] << endl;
}
}
// Function that pre-computes the dp array
void pre_compute(char a[][n])
{
// Check fo the first character
if (a[0][0] == 'a')
dp[0][0] = 0;
else
dp[0][0] = 1;
// Iterate in row and columns
for (int row = 0; row < n; row++) {
for (int col = 0; col < n; col++) {
// If not first row or not first column
if (row != 0 || col != 0)
dp[row][col] = INT_MAX;
// Not first row
if (row != 0) {
dp[row][col] = min(dp[row][col],
dp[row - 1][col]);
}
// Not first column
if (col != 0) {
dp[row][col] = min(dp[row][col],
dp[row][col - 1]);
}
// If it is not 'a' then increase by 1
if (a[row][col] != 'a' && (row != 0 || col != 0))
dp[row][col] += 1;
}
}
}
// Driver code
int main()
{
// character N X N array
char a[][3] = { { 'a', 'b', 'a' },
{ 'a', 'c', 'd' },
{ 'b', 'a', 'b' } };
// queries
pair queries[] = { { 1, 3 }, { 3, 3 } };
// number of queries
int q = 2;
// function call to pre-compute
pre_compute(a);
// function call to answer every query
answerQueries(queries, q);
} Java
// Java program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static int n = 3;
static int [][]dp = new int[n][n];
// Function to answer queries
static void answerQueries(pair queries[], int q)
{
// Iterate till query
for (int i = 0; i < q; i++)
{
// Decrease to get 0-based indexing
int x = queries[i].first;
x--;
int y = queries[i].second;
y--;
// Print answer
System.out.println(dp[x][y]);
}
}
// Function that pre-computes the dp array
static void pre_compute(char a[][])
{
// Check fo the first character
if (a[0][0] == 'a')
dp[0][0] = 0;
else
dp[0][0] = 1;
// Iterate in row and columns
for (int row = 0; row < n; row++)
{
for (int col = 0; col < n; col++)
{
// If not first row or not first column
if (row != 0 || col != 0)
dp[row][col] = Integer.MAX_VALUE;
// Not first row
if (row != 0)
{
dp[row][col] = Math.min(dp[row][col],
dp[row - 1][col]);
}
// Not first column
if (col != 0)
{
dp[row][col] = Math.min(dp[row][col],
dp[row][col - 1]);
}
// If it is not 'a' then increase by 1
if (a[row][col] != 'a' && (row != 0 || col != 0))
dp[row][col] += 1;
}
}
}
// Driver code
public static void main(String[] args)
{
// character N X N array
char a[][] = {{ 'a', 'b', 'a' },
{ 'a', 'c', 'd' },
{ 'b', 'a', 'b' }};
// queries
pair queries[] = { new pair( 1, 3 ),
new pair(3, 3 ) };
// number of queries
int q = 2;
// function call to pre-compute
pre_compute(a);
// function call to answer every query
answerQueries(queries, q);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find paths with maximum number
# of 'a' from (1, 1) to (X, Y) vertically
# or horizontally
n = 3
dp = [[0 for i in range(n)]
for j in range(n)]
# Function to answer queries
def answerQueries(queries, q):
# Iterate till query
for i in range(q):
# Decrease to get 0-based indexing
x = queries[i][0]
x -= 1
y = queries[i][1]
y -= 1
# Print answer
print(dp[x][y])
# Function that pre-computes the dp array
def pre_compute(a):
# Check for the first character
if a[0][0] == 'a':
dp[0][0] = 0
else:
dp[0][0] = 1
# Iterate in row and columns
for row in range(n):
for col in range(n):
# If not first row or not first column
if (row != 0 or col != 0):
dp[row][col] = 9999
# Not first row
if (row != 0):
dp[row][col] = min(dp[row][col],
dp[row - 1][col])
# Not first column
if (col != 0):
dp[row][col] = min(dp[row][col],
dp[row][col - 1])
# If it is not 'a' then increase by 1
if (a[row][col] != 'a' and (row != 0 or
col != 0)):
dp[row][col] += 1
# Driver code
if __name__ == '__main__':
# Character N X N array
a = [ ('a', 'b', 'a'),
('a', 'c', 'd'),
('b', 'a', 'b') ]
# Queries
queries = [ (1, 3), (3, 3) ]
# Number of queries
q = 2
# Function call to pre-compute
pre_compute(a)
# function call to answer every query
answerQueries(queries, q)
# This code is contributed by kirtishsurangalikarC#
// C# program to find paths with maximum number
// of 'a' from (1, 1) to (X, Y) vertically
// or horizontally
using System;
class GFG
{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
static int n = 3;
static int [,]dp = new int[n, n];
// Function to answer queries
static void answerQueries(pair []queries, int q)
{
// Iterate till query
for (int i = 0; i < q; i++)
{
// Decrease to get 0-based indexing
int x = queries[i].first;
x--;
int y = queries[i].second;
y--;
// Print answer
Console.WriteLine(dp[x, y]);
}
}
// Function that pre-computes the dp array
static void pre_compute(char [,]a)
{
// Check fo the first character
if (a[0, 0] == 'a')
dp[0, 0] = 0;
else
dp[0, 0] = 1;
// Iterate in row and columns
for (int row = 0; row < n; row++)
{
for (int col = 0; col < n; col++)
{
// If not first row or not first column
if (row != 0 || col != 0)
dp[row, col] = int.MaxValue;
// Not first row
if (row != 0)
{
dp[row, col] = Math.Min(dp[row, col],
dp[row - 1, col]);
}
// Not first column
if (col != 0)
{
dp[row, col] = Math.Min(dp[row, col],
dp[row, col - 1]);
}
// If it is not 'a' then increase by 1
if (a[row, col] != 'a' &&
(row != 0 || col != 0))
dp[row, col] += 1;
}
}
}
// Driver code
public static void Main(String[] args)
{
// character N X N array
char [,]a = {{ 'a', 'b', 'a' },
{ 'a', 'c', 'd' },
{ 'b', 'a', 'b' }};
// queries
pair []queries = { new pair(1, 3),
new pair(3, 3) };
// number of queries
int q = 2;
// function call to pre-compute
pre_compute(a);
// function call to answer every query
answerQueries(queries, q);
}
}
// This code is contributed by PrinciRaj1992Javascript
输出
1
2时间复杂度：O(N 2 ) 用于预计算，O(1) 用于每个查询。
辅助空间：O(N 2 )
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