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Jun 22, 2024
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feat: add solutions to lc problem: No.2101 #3147

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1 change: 0 additions & 1 deletion .prettierignore
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Expand Up @@ -15,7 +15,6 @@ node_modules/
/solution/bash_problem_readme_template_en.md
/solution/0100-0199/0177.Nth Highest Salary/Solution.sql
/solution/0100-0199/0178.Rank Scores/Solution2.sql
/solution/0100-0199/0196.Delete Duplicate Emails/Solution3.sql
/solution/0500-0599/0586.Customer Placing the Largest Number of Orders/Solution2.sql
/solution/1400-1499/1454.Active Users/Solution.sql
/solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
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1 change: 1 addition & 0 deletions .prettierrc
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Expand Up @@ -11,6 +11,7 @@
"braceStyle": "1tbs",
"endOfLine": "lf",
"sqlKeywordCase": "upper",
"sqlCanonicalSyntax": false,
"overrides": [
{
"files": ["*.sql"],
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214 changes: 111 additions & 103 deletions solution/2100-2199/2101.Detonate the Maximum Bombs/README.md
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Expand Up @@ -88,7 +88,13 @@ tags:

### 方法一：BFS

枚举每个炸弹 k 作为起始引爆点，BFS 搜索能影响到的所有炸弹的数量，取其最大值。
我们定义一个长度为 $n$ 的数组 $g$，其中 $g[i]$ 表示炸弹 $i$ 的爆炸范围内可以引爆的所有炸弹的下标。

然后，我们遍历所有炸弹，对于两个炸弹 $(x_1, y_1, r_1)$ 和 $(x_2, y_2, r_2)$，我们计算它们之间的距离 $\text{dist} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$。如果 $\text{dist} \leq r_1$，那么炸弹 $i$ 的爆炸范围内可以引爆炸弹 $j$，我们就将 $j$ 添加到 $g[i]$ 中。如果 $\text{dist} \leq r_2$，那么炸弹 $j$ 的爆炸范围内可以引爆炸弹 $i$，我们就将 $i$ 添加到 $g[j]$ 中。

接下来，我们遍历所有炸弹，对于每个炸弹 $k$，我们使用广度优先搜索计算炸弹 $k$ 的爆炸范围内可以引爆的所有炸弹的下标，并记录下来。如果这些炸弹的数量等于 $n$，那么我们就可以引爆所有炸弹，直接返回 $n$。否则，我们记录下来这些炸弹的数量，并返回最大值。

时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为炸弹的数量。

<!-- tabs:start -->

Expand All @@ -97,82 +103,77 @@ tags:
```python
class Solution:
def maximumDetonation(self, bombs: List[List[int]]) -> int:
def check(i, j):
if i == j:
return False
x, y = bombs[i][0] - bombs[j][0], bombs[i][1] - bombs[j][1]
r = bombs[i][2]
return r * r >= x * x + y * y

g = defaultdict(list)
n = len(bombs)
for i in range(n):
for j in range(n):
if check(i, j):
g = [[] for _ in range(n)]
for i in range(n - 1):
x1, y1, r1 = bombs[i]
for j in range(i + 1, n):
x2, y2, r2 = bombs[j]
dist = hypot(x1 - x2, y1 - y2)
if dist <= r1:
g[i].append(j)
if dist <= r2:
g[j].append(i)
ans = 0
for k in range(n):
q = deque([k])
vis = [False] * n
vis[k] = True
cnt = 0
while q:
i = q.popleft()
cnt += 1
vis = {k}
q = [k]
for i in q:
for j in g[i]:
if not vis[j]:
vis[j] = True
if j not in vis:
vis.add(j)
q.append(j)
ans = max(ans, cnt)
if len(vis) == n:
return n
ans = max(ans, len(vis))
return ans
```

#### Java

```java
class Solution {
private int[][] bombs;

public int maximumDetonation(int[][] bombs) {
this.bombs = bombs;
int n = bombs.length;
boolean[][] g = new boolean[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
g[i][j] = check(i, j);
List<Integer>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
int[] p1 = bombs[i], p2 = bombs[j];
double dist = Math.hypot(p1[0] - p2[0], p1[1] - p2[1]);
if (dist <= p1[2]) {
g[i].add(j);
}
if (dist <= p2[2]) {
g[j].add(i);
}
}
}
int ans = 0;
boolean[] vis = new boolean[n];
for (int k = 0; k < n; ++k) {
Deque<Integer> q = new ArrayDeque<>();
q.offer(k);
boolean[] vis = new boolean[n];
Arrays.fill(vis, false);
vis[k] = true;
int cnt = 0;
Deque<Integer> q = new ArrayDeque<>();
q.offer(k);
while (!q.isEmpty()) {
int i = q.poll();
++cnt;
for (int j = 0; j < n; ++j) {
if (g[i][j] && !vis[j]) {
for (int j : g[i]) {
if (!vis[j]) {
vis[j] = true;
q.offer(j);
}
}
}
if (cnt == n) {
return n;
}
ans = Math.max(ans, cnt);
}
return ans;
}

private boolean check(int i, int j) {
if (i == j) {
return false;
}
long x = bombs[i][0] - bombs[j][0];
long y = bombs[i][1] - bombs[j][1];
long r = bombs[i][2];
return r * r >= x * x + y * y;
}
}
```

Expand All @@ -183,64 +184,67 @@ class Solution {
public:
int maximumDetonation(vector<vector<int>>& bombs) {
int n = bombs.size();
vector<vector<bool>> g(n, vector<bool>(n));
for (int i = 0; i < n; ++i)
for (int j = 0; j < n; ++j)
g[i][j] = check(i, j, bombs);
vector<int> g[n];
for (int i = 0; i < n - 1; ++i) {
for (int j = i + 1; j < n; ++j) {
auto& p1 = bombs[i];
auto& p2 = bombs[j];
auto dist = hypot(p1[0] - p2[0], p1[1] - p2[1]);
if (dist <= p1[2]) {
g[i].push_back(j);
}
if (dist <= p2[2]) {
g[j].push_back(i);
}
}
}
int ans = 0;
bool vis[n];
for (int k = 0; k < n; ++k) {
queue<int> q{{k}};
vector<bool> vis(n);
memset(vis, false, sizeof(vis));
queue<int> q;
q.push(k);
vis[k] = true;
int cnt = 0;
while (!q.empty()) {
int i = q.front();
q.pop();
++cnt;
for (int j = 0; j < n; ++j) {
if (g[i][j] && !vis[j]) {
for (int j : g[i]) {
if (!vis[j]) {
vis[j] = true;
q.push(j);
}
}
}
if (cnt == n) {
return n;
}
ans = max(ans, cnt);
}
return ans;
}

bool check(int i, int j, vector<vector<int>>& bombs) {
if (i == j) return false;
long long x = bombs[i][0] - bombs[j][0];
long long y = bombs[i][1] - bombs[j][1];
long long r = bombs[i][2];
return r * r >= x * x + y * y;
}
};
```

#### Go

```go
func maximumDetonation(bombs [][]int) int {
check := func(i, j int) bool {
if i == j {
return false
}
x, y := bombs[i][0]-bombs[j][0], bombs[i][1]-bombs[j][1]
r := bombs[i][2]
return r*r >= x*x+y*y
}
func maximumDetonation(bombs [][]int) (ans int) {
n := len(bombs)
g := make([][]bool, n)
for i := range g {
g[i] = make([]bool, n)
for j := range g[i] {
g[i][j] = check(i, j)
g := make([][]int, n)
for i, p1 := range bombs[:n-1] {
for j := i + 1; j < n; j++ {
p2 := bombs[j]
dist := math.Hypot(float64(p1[0]-p2[0]), float64(p1[1]-p2[1]))
if dist <= float64(p1[2]) {
g[i] = append(g[i], j)
}
if dist <= float64(p2[2]) {
g[j] = append(g[j], i)
}
}
}

ans := 0
for k := 0; k < n; k++ {
q := []int{k}
vis := make([]bool, n)
Expand All @@ -250,16 +254,19 @@ func maximumDetonation(bombs [][]int) int {
i := q[0]
q = q[1:]
cnt++
for j := 0; j < n; j++ {
if g[i][j] && !vis[j] {
for _, j := range g[i] {
if !vis[j] {
vis[j] = true
q = append(q, j)
}
}
}
if cnt == n {
return n
}
ans = max(ans, cnt)
}
return ans
return
}
```

Expand All @@ -268,37 +275,38 @@ func maximumDetonation(bombs [][]int) int {
```ts
function maximumDetonation(bombs: number[][]): number {
const n = bombs.length;
const g = new Map<number, number[]>(bombs.map((_, i) => [i, []]));

for (let i = 0; i < n - 1; i++) {
for (let j = 1; j < n; j++) {
const [x1, y1, r1] = bombs[i];
const g: number[][] = Array.from({ length: n }, () => []);
for (let i = 0; i < n - 1; ++i) {
const [x1, y1, r1] = bombs[i];
for (let j = i + 1; j < n; ++j) {
const [x2, y2, r2] = bombs[j];
const distance = Math.hypot(x1 - x2, y1 - y2);

if (distance <= r1) g.get(i)!.push(j);
if (distance <= r2) g.get(j)!.push(i);
const d = Math.hypot(x1 - x2, y1 - y2);
if (d <= r1) {
g[i].push(j);
}
if (d <= r2) {
g[j].push(i);
}
}
}

let res = 0;
for (let i = 0; i < n; i++) {
const seen = new Set<number>([i]);
const q = [i];

let ans = 0;
for (let k = 0; k < n; ++k) {
const vis: Set<number> = new Set([k]);
const q: number[] = [k];
for (const i of q) {
for (const j of g.get(i) ?? []) {
if (seen.has(j)) continue;
seen.add(j);
q.push(j);
for (const j of g[i]) {
if (!vis.has(j)) {
vis.add(j);
q.push(j);
}
}
}

if (seen.size === n) return n;
res = Math.max(res, seen.size);
if (vis.size === n) {
return n;
}
ans = Math.max(ans, vis.size);
}

return res;
return ans;
}
```

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