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Merge pull request #919 from 0xff-dev/2285
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Add solution and test-cases for problem 2285
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6boris authored Jun 30, 2024
2 parents 79e6ba4 + a52438b commit 7631850
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46 changes: 32 additions & 14 deletions leetcode/2201-2300/2285.Maximum-Total-Importance-of-Roads/README.md
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# [2285.Maximum Total Importance of Roads][title]

> [!WARNING|style:flat]
> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
## Description
You are given an integer `n` denoting the number of cities in a country. The cities are numbered from `0` to `n - 1`.

You are also given a 2D integer array `roads` where `roads[i] = [ai, bi]` denotes that there exists a **bidirectional** road connecting cities `ai` and `bi`.

You need to assign each city with an integer value from `1` to `n`, where each value can only be used **once**. The **importance** of a road is then defined as the **sum** of the values of the two cities it connects.

Return the **maximum total importance** of all roads possible after assigning the values optimally.

**Example 1:**
**Example 1:**

![1](./ex1drawio.png)

```
Input: a = "11", b = "1"
Output: "100"
Input: n = 5, roads = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]]
Output: 43
Explanation: The figure above shows the country and the assigned values of [2,4,5,3,1].
- The road (0,1) has an importance of 2 + 4 = 6.
- The road (1,2) has an importance of 4 + 5 = 9.
- The road (2,3) has an importance of 5 + 3 = 8.
- The road (0,2) has an importance of 2 + 5 = 7.
- The road (1,3) has an importance of 4 + 3 = 7.
- The road (2,4) has an importance of 5 + 1 = 6.
The total importance of all roads is 6 + 9 + 8 + 7 + 7 + 6 = 43.
It can be shown that we cannot obtain a greater total importance than 43.
```

## 题意
> ...
**Example 2:**

## 题解
![2](./ex2drawio.png)

### 思路1
> ...
Maximum Total Importance of Roads
```go
```

Input: n = 5, roads = [[0,3],[2,4],[1,3]]
Output: 20
Explanation: The figure above shows the country and the assigned values of [4,3,2,5,1].
- The road (0,3) has an importance of 4 + 5 = 9.
- The road (2,4) has an importance of 2 + 1 = 3.
- The road (1,3) has an importance of 3 + 5 = 8.
The total importance of all roads is 9 + 3 + 8 = 20.
It can be shown that we cannot obtain a greater total importance than 20.
```

## 结语

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package Solution

func Solution(x bool) bool {
return x
import "sort"

func Solution(n int, roads [][]int) int64 {
ans := int64(0)

// 检查入读和出度
in := make([]int, n)
for _, r := range roads {
in[r[0]]++
in[r[1]]++
}
sort.Ints(in)
v := int64(1)
for _, i := range in {
ans += v * int64(i)
v++
}
return ans
}
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Expand Up @@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
// 测试用例
cases := []struct {
name string
inputs bool
expect bool
n int
roads [][]int
expect int64
}{
{"TestCase", true, true},
{"TestCase", true, true},
{"TestCase", false, false},
{"TestCase1", 5, [][]int{{0, 1}, {1, 2}, {2, 3}, {0, 2}, {1, 3}, {2, 4}}, 43},
{"TestCase2", 5, [][]int{{0, 3}, {2, 4}, {1, 3}}, 20},
}

// 开始测试
for i, c := range cases {
t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
got := Solution(c.inputs)
got := Solution(c.n, c.roads)
if !reflect.DeepEqual(got, c.expect) {
t.Fatalf("expected: %v, but got: %v, with inputs: %v",
c.expect, got, c.inputs)
t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
c.expect, got, c.n, c.roads)
}
})
}
}

// 压力测试
// 压力测试
func BenchmarkSolution(b *testing.B) {
}

// 使用案列
// 使用案列
func ExampleSolution() {
}
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