2022-06-20：一个二维矩阵，上面只有 0 和 1，只能上下左右移动，
如果移动前后的元素值相同，则耗费 1 ，否则耗费 2。
问从左上到右下的最小耗费。
来自网易。3.27笔试。
答案2022-06-20：
1.网上非常流行的方法，但这是错误的。这道题动态规划是做不了的。因为上下左右四个方向都可能走，而不是右下两个方向。
2.要用dijskra 小根堆才能实现。
代码里1和2两种方法都实现了，运行结果可以证明方法1是错误的。
代码用rust编写。代码如下：
代码语言：rust复制use rand::Rng;
fn main() {
let n: i32 = 100;
let m: i32 = 100;
let test_time: i32 = 10000;
println!("测试开始");
for i in 0..test_time {
let mut map = random_matrix(n, m);
let ans1 = best_walk1(&mut map);
let ans2 = best_walk2(&mut map);
if ans1 != ans2 {
println!("出错了!{}", i);
println!("ans1 = {}", ans1);
println!("ans2 = {}", ans2);
break;
}
}
println!("测试结束");
}
// 一个错误的贪心
// 网上帖子最流行的解答，看似对，其实不行
fn best_walk1(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 i in 0..n {
dp.push(vec![]);
for _ in 0..m {
dp[i as usize].push(0);
}
}
for i in 1..m {
dp[0][i as usize] = dp[0][(i - 1) as usize]
if map[0][(i - 1) as usize] == map[0][i as usize] {
1
} else {
2
};
}
for i in 1..n {
dp[i as usize][0] = dp[(i - 1) as usize][0]
if map[(i - 1) as usize][0] == map[i as usize][0] {
1
} else {
2
};
}
for i in 1..n {
for j in 1..m {
dp[i as usize][j as usize] = dp[(i - 1) as usize][j as usize]
(if map[(i - 1) as usize][j as usize] == map[i as usize][j as usize] {
1
} else {
2
});
dp[i as usize][j as usize] = get_min(
dp[i as usize][j as usize],
dp[i as usize][(j - 1) as usize]
(if map[i as usize][(j - 1) as usize] == map[i as usize][j as usize] {
1
} else {
2
}),
);
}
}
return dp[(n - 1) as usize][(m - 1) as usize];
}
fn get_min<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T {
if a < b {
a
} else {
b
}
}
// 正确的解法
// Dijskra
fn best_walk2(map: &mut Vec<Vec<i32>>) -> i32 {
let n = map.len() as i32;
let m = map[0].len() as i32;
// 小根堆：[代价，行，列]
// 根据代价，谁代价小，谁放在堆的上面
let mut heap: Vec<Vec<i32>> = vec![];
// poped[i][j] == true 已经弹出过了！不要再处理，直接忽略！
// poped[i][j] == false 之间(i,j)没弹出过！要处理
let mut poped: Vec<Vec<bool>> = vec![];
for i in 0..n {
poped.push(vec![]);
for _ in 0..m {
poped[i as usize].push(false);
}
}
heap.push(vec![0, 0, 0]);
let mut ans = 0;
while heap.len() > 0 {
// 当前弹出了，[代价，行，列]，当前位置
heap.sort_by(|a, b| b[0].cmp(&a[0]));
let cur = heap.pop().unwrap();
let dis = cur[0];
let row = cur[1];
let col = cur[2];
if poped[row as usize][col as usize] {
continue;
}
// 第一次弹出！
poped[row as usize][col as usize] = true;
if row == n - 1 && col == m - 1 {
ans = dis;
break;
}
add(
dis,
row - 1,
col,
map[row as usize][col as usize],
n,
m,
map,
&mut poped,
&mut heap,
);
add(
dis,
row 1,
col,
map[row as usize][col as usize],
n,
m,
map,
&mut poped,
&mut heap,
);
add(
dis,
row,
col - 1,
map[row as usize][col as usize],
n,
m,
map,
&mut poped,
&mut heap,
);
add(
dis,
row,
col 1,
map[row as usize][col as usize],
n,
m,
map,
&mut poped,
&mut heap,
);
}
return ans;
}
// preDistance ： 之前的距离
// int row, int col ： 当前要加入的是什么位置
// preValue : 前一个格子是什么值，
// int n, int m ：边界，固定参数
// map: 每一个格子的值，都在map里
// boolean[][] poped : 当前位置如果是弹出过的位置，要忽略！
// PriorityQueue<int[]> heap : 小根堆
fn add(
pre_distance: i32,
row: i32,
col: i32,
pre_value: i32,
n: i32,
m: i32,
map: &mut Vec<Vec<i32>>,
poped: &mut Vec<Vec<bool>>,
heap: &mut Vec<Vec<i32>>,
) {
if row >= 0 && row < n && col >= 0 && col < m && !poped[row as usize][col as usize] {
heap.push(vec![
pre_distance
if map[row as usize][col as usize] == pre_value {
1
} else {
2
},
row,
col,
]);
}
}
// 为了测试
fn random_matrix(n: i32, m: i32) -> Vec<Vec<i32>> {
let mut ans: Vec<Vec<i32>> = vec![];
for i in 0..n {
ans.push(vec![]);
for _ in 0..m {
ans[i as usize].push(rand::thread_rng().gen_range(0, 2));
}
}
return ans;
}执行结果如下：
在这里插入图片描述左神java代码
