2022-06-10：薯队长从北向南穿过一片红薯地（南北长M，东西宽N），红薯地被划分为1x1的方格，
他可以从北边的任何一个格子出发，到达南边的任何一个格子，
但每一步只能走到东南、正南、西南方向的三个格子之一，
而且不能跨出红薯地，他可以获得经过的格子上的所有红薯，请问他可以获得最多的红薯个数。
来自小红书，小红书第一题。
答案2022-06-10：
动态规划。dp是两行格子。dp[0]是加arr[i][j]之前的最大值数组。dp[1]是加arr[i][j]之后最大值数组。
dp[1][j]=arr[i][j]+max(dp[0][j-1],dp[0][j],dp[[0][j+1])。未来不确定，但是过去是确定的。dp[0]代表过去，dp[1]根据过去的三条方向选择最优方向即可。
时间复杂度：O(MN)。
空间复杂度：O(N)。占用两行格子。
代码用rust编写。代码如下：
fn main() {
let mut map: Vec<Vec<i32>> = vec![
vec![3, 5, 6, 2, 4],
vec![7, 8, 9, 1, 1],
vec![1, 2, 3, 4, 5],
];
let ans = max_score(&mut map);
println!("ans = {}", ans);
let mut map: Vec<Vec<i32>> = vec![
vec![3, 5, 6, 2, 4],
vec![7, 8, 9, 1, 1],
vec![1, 2, 3, 4, 5],
];
let ans = max_score2(&mut map);
println!("ans = {}", ans);
let mut map: Vec<Vec<i32>> = vec![
vec![3, 5, 6, 2, 4],
vec![7, 8, 9, 1, 1],
vec![1, 2, 3, 4, 5],
];
let ans = max_score3(&mut map);
println!("ans = {}", ans);
}
fn max_score(map: &mut Vec<Vec<i32>>) -> i32 {
let mut ans = 0;
for col in 0..map[0].len() as i32 {
ans = get_max(ans, process(map, 0, col));
}
return ans;
}
fn process(map: &mut Vec<Vec<i32>>, row: i32, col: i32) -> i32 {
if col < 0 || col == map[0].len() as i32 {
return -1;
}
if row == map.len() as i32 - 1 {
return map[row as usize][col as usize];
}
let cur = map[row as usize][col as usize];
let next1 = process(map, row + 1, col - 1);
let next2 = process(map, row + 1, col);
let next3 = process(map, row + 1, col + 1);
return cur + get_max(get_max(next1, next2), next3);
}
fn max_score2(map: &mut Vec<Vec<i32>>) -> i32 {
let mut ans = 0;
let n = map.len() as i32;
let m = map[0].len() as i32;
let mut dp: Vec<Vec<i32>> = vec![];
for i in 0..n {
dp.push(vec![]);
for _ in 0..m {
dp[i as usize].push(-2);
}
}
for col in 0..map[0].len() as i32 {
ans = get_max(ans, process2(map, 0, col, &mut dp));
}
return ans;
}
fn process2(map: &mut Vec<Vec<i32>>, row: i32, col: i32, dp: &mut Vec<Vec<i32>>) -> i32 {
if col < 0 || col == map[0].len() as i32 {
return -1;
}
if dp[row as usize][col as usize] != -2 {
return dp[row as usize][col as usize];
}
// 继续算！
let mut ans = 0;
if row == map.len() as i32 - 1 {
ans = map[row as usize][col as usize];
} else {
let cur = map[row as usize][col as usize];
let next1 = process2(map, row + 1, col - 1, dp);
let next2 = process2(map, row + 1, col, dp);
let next3 = process2(map, row + 1, col + 1, dp);
ans = cur + get_max(get_max(next1, next2), next3);
}
dp[row as usize][col as usize] = ans;
return ans;
}
// 最优方法
fn max_score3(map: &mut Vec<Vec<i32>>) -> i32 {
let n = map.len() as i32;
let m = map[0].len() as i32;
let mut dp: Vec<Vec<i32>> = vec![];
for _ in 0..2 {
dp.push(map[0].clone());
}
for i in 1..n {
for j in 0..m as i32 {
dp[1][j as usize] = get_max(
get_max(
dp[0][j as usize],
if j - 1 >= 0 {
dp[0][(j - 1) as usize]
} else {
0
},
),
if j + 1 < m {
dp[0][(j + 1) as usize]
} else {
0
},
) + map[i as usize][j as usize];
}
dp[0] = dp[1].clone();
}
let mut ans = dp[0][0];
for i in 1..m {
ans = get_max(ans, dp[0][i as usize]);
}
return ans;
}
fn get_max<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T {
if a > b {
a
} else {
b
}
}
执行结果如下：
左神java代码