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feat: add solutions to lc problem: No.1928 #3597

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129 changes: 125 additions & 4 deletions solution/1900-1999/1928.Minimum Cost to Reach Destination in Time/README.md
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Expand Up @@ -81,32 +81,153 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：动态规划

我们定义 $f[i][j]$ 表示经过 $i$ 分钟，从城市 $0$ 到达城市 $j$ 的最小花费。初始时 $f[0][0] = \textit{passingFees}[0]$，其余的 $f[0][j] = +\infty$。

接下来，我们在 $[1, \textit{maxTime}]$ 的时间范围内，遍历所有的边，对于每一条边 $(x, y, t)$，如果 $t \leq i$，那么我们：

- 可以先经过 $i - t$ 分钟，从城市 $0$ 到达城市 $y$，然后再经过 $t$ 分钟，从城市 $y$ 到达城市 $x$，再加上到达城市 $x$ 的通行费，即 $f[i][x] = \min(f[i][x], f[i - t][y] + \textit{passingFees}[x])$；
- 也可以先经过 $i - t$ 分钟，从城市 $0$ 到达城市 $x$，然后再经过 $t$ 分钟，从城市 $x$ 到达城市 $y$，再加上到达城市 $y$ 的通行费，即 $f[i][y] = \min(f[i][y], f[i - t][x] + \textit{passingFees}[y])$。

最终答案即为 $\min\{f[i][n - 1]\}$，其中 $i \in [0, \textit{maxTime}]$。如果答案为 $+\infty$，则返回 $-1$。

时间复杂度 $O(\textit{maxTime} \times (m + n))$，其中 $m$ 和 $n$ 分别是边的数量和城市的数量。空间复杂度 $O(\textit{maxTime} \times n)$。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def minCost(
self, maxTime: int, edges: List[List[int]], passingFees: List[int]
) -> int:
m, n = maxTime, len(passingFees)
f = [[inf] * n for _ in range(m + 1)]
f[0][0] = passingFees[0]
for i in range(1, m + 1):
for x, y, t in edges:
if t <= i:
f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x])
f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y])
ans = min(f[i][n - 1] for i in range(m + 1))
return ans if ans < inf else -1
```

#### Java

```java

class Solution {
public int minCost(int maxTime, int[][] edges, int[] passingFees) {
int m = maxTime, n = passingFees.length;
int[][] f = new int[m + 1][n];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, inf);
}
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (var e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = Math.min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = Math.min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 0; i <= m; ++i) {
ans = Math.min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int minCost(int maxTime, vector<vector<int>>& edges, vector<int>& passingFees) {
int m = maxTime, n = passingFees.size();
const int inf = 1 << 30;
vector<vector<int>> f(m + 1, vector<int>(n, inf));
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (const auto& e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 1; i <= m; ++i) {
ans = min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
};
```

#### Go

```go
func minCost(maxTime int, edges [][]int, passingFees []int) int {
m, n := maxTime, len(passingFees)
f := make([][]int, m+1)
const inf int = 1 << 30
for i := range f {
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = inf
}
}
f[0][0] = passingFees[0]
for i := 1; i <= m; i++ {
for _, e := range edges {
x, y, t := e[0], e[1], e[2]
if t <= i {
f[i][x] = min(f[i][x], f[i-t][y]+passingFees[x])
f[i][y] = min(f[i][y], f[i-t][x]+passingFees[y])
}
}
}
ans := inf
for i := 1; i <= m; i++ {
ans = min(ans, f[i][n-1])
}
if ans == inf {
return -1
}
return ans
}
```

#### TypeScript

```ts
function minCost(maxTime: number, edges: number[][], passingFees: number[]): number {
const [m, n] = [maxTime, passingFees.length];
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n).fill(Infinity));
f[0][0] = passingFees[0];
for (let i = 1; i <= m; ++i) {
for (const [x, y, t] of edges) {
if (t <= i) {
f[i][x] = Math.min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = Math.min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
let ans = Infinity;
for (let i = 1; i <= m; ++i) {
ans = Math.min(ans, f[i][n - 1]);
}
return ans === Infinity ? -1 : ans;
}
```

<!-- tabs:end -->
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129 changes: 125 additions & 4 deletions solution/1900-1999/1928.Minimum Cost to Reach Destination in Time/README_EN.md
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Expand Up @@ -79,32 +79,153 @@ You cannot take path 0 -&gt; 1 -&gt; 2 -&gt; 5 since it would take too long.

<!-- solution:start -->

### Solution 1
### Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the minimum cost to reach city $j$ from city $0$ after $i$ minutes. Initially, $f[0][0] = \textit{passingFees}[0]$, and the rest $f[0][j] = +\infty$.

Next, within the time range $[1, \textit{maxTime}]$, we traverse all edges. For each edge $(x, y, t)$, if $t \leq i$, then we:

- Can first spend $i - t$ minutes to reach city $y$ from city $0$, then spend $t$ minutes to reach city $x$ from city $y$, and add the passing fee to reach city $x$, i.e., $f[i][x] = \min(f[i][x], f[i - t][y] + \textit{passingFees}[x])$;
- Can also first spend $i - t$ minutes to reach city $x$ from city $0$, then spend $t$ minutes to reach city $y$ from city $x$, and add the passing fee to reach city $y$, i.e., $f[i][y] = \min(f[i][y], f[i - t][x] + \textit{passingFees}[y])$.

The final answer is $\min\{f[i][n - 1]\}$, where $i \in [0, \textit{maxTime}]$. If the answer is $+\infty$, return $-1$.

The time complexity is $O(\textit{maxTime} \times (m + n))$, where $m$ and $n$ are the number of edges and cities, respectively. The space complexity is $O(\textit{maxTime} \times n)$.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def minCost(
self, maxTime: int, edges: List[List[int]], passingFees: List[int]
) -> int:
m, n = maxTime, len(passingFees)
f = [[inf] * n for _ in range(m + 1)]
f[0][0] = passingFees[0]
for i in range(1, m + 1):
for x, y, t in edges:
if t <= i:
f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x])
f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y])
ans = min(f[i][n - 1] for i in range(m + 1))
return ans if ans < inf else -1
```

#### Java

```java

class Solution {
public int minCost(int maxTime, int[][] edges, int[] passingFees) {
int m = maxTime, n = passingFees.length;
int[][] f = new int[m + 1][n];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, inf);
}
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (var e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = Math.min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = Math.min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 0; i <= m; ++i) {
ans = Math.min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int minCost(int maxTime, vector<vector<int>>& edges, vector<int>& passingFees) {
int m = maxTime, n = passingFees.size();
const int inf = 1 << 30;
vector<vector<int>> f(m + 1, vector<int>(n, inf));
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (const auto& e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 1; i <= m; ++i) {
ans = min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
};
```

#### Go

```go
func minCost(maxTime int, edges [][]int, passingFees []int) int {
m, n := maxTime, len(passingFees)
f := make([][]int, m+1)
const inf int = 1 << 30
for i := range f {
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = inf
}
}
f[0][0] = passingFees[0]
for i := 1; i <= m; i++ {
for _, e := range edges {
x, y, t := e[0], e[1], e[2]
if t <= i {
f[i][x] = min(f[i][x], f[i-t][y]+passingFees[x])
f[i][y] = min(f[i][y], f[i-t][x]+passingFees[y])
}
}
}
ans := inf
for i := 1; i <= m; i++ {
ans = min(ans, f[i][n-1])
}
if ans == inf {
return -1
}
return ans
}
```

#### TypeScript

```ts
function minCost(maxTime: number, edges: number[][], passingFees: number[]): number {
const [m, n] = [maxTime, passingFees.length];
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n).fill(Infinity));
f[0][0] = passingFees[0];
for (let i = 1; i <= m; ++i) {
for (const [x, y, t] of edges) {
if (t <= i) {
f[i][x] = Math.min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = Math.min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
let ans = Infinity;
for (let i = 1; i <= m; ++i) {
ans = Math.min(ans, f[i][n - 1]);
}
return ans === Infinity ? -1 : ans;
}
```

<!-- tabs:end -->
Expand Down
23 changes: 23 additions & 0 deletions solution/1900-1999/1928.Minimum Cost to Reach Destination in Time/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
class Solution {
public:
int minCost(int maxTime, vector<vector<int>>& edges, vector<int>& passingFees) {
int m = maxTime, n = passingFees.size();
const int inf = 1 << 30;
vector<vector<int>> f(m + 1, vector<int>(n, inf));
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (const auto& e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 1; i <= m; ++i) {
ans = min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
};
29 changes: 29 additions & 0 deletions solution/1900-1999/1928.Minimum Cost to Reach Destination in Time/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,29 @@
func minCost(maxTime int, edges [][]int, passingFees []int) int {
m, n := maxTime, len(passingFees)
f := make([][]int, m+1)
const inf int = 1 << 30
for i := range f {
f[i] = make([]int, n)
for j := range f[i] {
f[i][j] = inf
}
}
f[0][0] = passingFees[0]
for i := 1; i <= m; i++ {
for _, e := range edges {
x, y, t := e[0], e[1], e[2]
if t <= i {
f[i][x] = min(f[i][x], f[i-t][y]+passingFees[x])
f[i][y] = min(f[i][y], f[i-t][x]+passingFees[y])
}
}
}
ans := inf
for i := 1; i <= m; i++ {
ans = min(ans, f[i][n-1])
}
if ans == inf {
return -1
}
return ans
}
25 changes: 25 additions & 0 deletions solution/1900-1999/1928.Minimum Cost to Reach Destination in Time/Solution.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,25 @@
class Solution {
public int minCost(int maxTime, int[][] edges, int[] passingFees) {
int m = maxTime, n = passingFees.length;
int[][] f = new int[m + 1][n];
final int inf = 1 << 30;
for (var g : f) {
Arrays.fill(g, inf);
}
f[0][0] = passingFees[0];
for (int i = 1; i <= m; ++i) {
for (var e : edges) {
int x = e[0], y = e[1], t = e[2];
if (t <= i) {
f[i][x] = Math.min(f[i][x], f[i - t][y] + passingFees[x]);
f[i][y] = Math.min(f[i][y], f[i - t][x] + passingFees[y]);
}
}
}
int ans = inf;
for (int i = 0; i <= m; ++i) {
ans = Math.min(ans, f[i][n - 1]);
}
return ans == inf ? -1 : ans;
}
}
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