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2435. 矩阵中和能被 K 整除的路径

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li-chx 2025-11-26 08:56:35 +08:00
parent 0babcf1472
commit 1da8b6b735
1 changed files with 17 additions and 37 deletions
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54
src/main.rs
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@ -1,50 +1,30 @@
use crate::arr::make_string_arr; use crate::arr::make_matrix;
use std::cmp::max;
struct Solution; struct Solution;
mod arr; mod arr;
impl Solution { impl Solution {
pub fn find_max_form(strs: Vec<String>, m: i32, n: i32) -> i32 { pub fn number_of_paths(grid: Vec<Vec<i32>>, k: i32) -> i32 {
let mut dp = vec![vec![0; m as usize + 1]; n as usize + 1]; let mut arr = vec![vec![vec![0; k as usize]; grid[0].len()]; grid.len()];
fn get_count(str: &String) -> (i32, i32) { let r#mod = 1e9 as i32 + 7;
let (mut zeros, mut ones) = (0, 0); arr[0][0][(grid[0][0] % k) as usize] = 1;
for c in str.chars() { for i in 0..grid.len() {
if c == '0' { for j in 0..grid[i].len() {
zeros += 1; if i < grid.len() - 1 {
} else { for l in 0..k {
ones += 1; arr[i + 1][j][((l + grid[i + 1][j]) % k) as usize] = (arr[i + 1][j][((l + grid[i + 1][j]) % k) as usize] + arr[i][j][l as usize]) % r#mod;
}
} }
} if j < grid[0].len() - 1 {
(zeros, ones) for l in 0..k {
} arr[i][j + 1][((l + grid[i][j + 1]) % k) as usize] = (arr[i][j + 1][((l + grid[i][j + 1]) % k) as usize] + arr[i][j][l as usize]) % r#mod;
let mut strs = strs }
.iter()
.map(|x| get_count(x))
.collect::<Vec<(i32, i32)>>();
// n1 m0
// 外1内0
for i in 0..strs.len() {
let (zeros, ones) = strs[i];
for i in (ones..=n).rev() {
for k in (zeros..=m).rev() {
dp[i as usize][k as usize] = dp[i as usize][k as usize]
.max(dp[(i - ones) as usize][(k - zeros) as usize] + 1);
} }
} }
} }
let mut ans = 0; arr[grid.len() - 1][grid[0].len() - 1][0]
dp.iter()
.for_each(|x| x.iter().for_each(|y| ans = max(ans, *y)));
ans
} }
} }
// dp?
fn main() { fn main() {
let result = Solution::find_max_form( let result = Solution::number_of_paths(make_matrix("[[5,2,4],[3,0,5],[0,7,2]]"), 3);
make_string_arr("[\"10\", \"0001\", \"111001\", \"1\", \"0\"]"),
5,
3,
);
println!("{:?}", result); println!("{:?}", result);
} }