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feat: add solutions to lc problem: No.0749
No.0749.Contain Virus
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‎README.md

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感谢以下所有朋友对本项目的贡献！
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<a href="https://github.com/doocs/leetcode/graphs/contributors" target="_blank"><img src="https://contrib.rocks/image?repo=doocs/leetcode&max=500" /></a>
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<a href="https://github.com/doocs/leetcode/graphs/contributors" target="_blank"><img src="https://contrib.rocks/image?repo=doocs/leetcode&max=499" /></a>
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## 赞助者
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‎README_EN.md

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This project exists thanks to all the people who contribute.
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<a href="https://github.com/doocs/leetcode/graphs/contributors" target="_blank"><img src="https://contrib.rocks/image?repo=doocs/leetcode&max=500" /></a>
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<a href="https://github.com/doocs/leetcode/graphs/contributors" target="_blank"><img src="https://contrib.rocks/image?repo=doocs/leetcode&max=499" /></a>
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## Backers & Sponsors
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‎solution/0700-0799/0749.Contain Virus/README.md

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<strong>输出:</strong> 10
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<strong>解释:</strong>一共有两块被病毒感染的区域。
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在第一天，添加 5 墙隔离病毒区域的左侧。病毒传播后的状态是:
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<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0749.Contain%20Virus/images/653" />
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<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0749.Contain%20Virus/images/virus12edited-grid.jpg" />
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第二天，在右侧添加 5 个墙来隔离病毒区域。此时病毒已经被完全控制住了。
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<img src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0749.Contain%20Virus/images/virus13edited-grid.jpg" style="height: 261px; width: 500px;" />
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</pre>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：DFS 暴力模拟**
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DFS 找到每个病毒区域 `areas[i]`，同时记录每个区域边界节点 `boundaries[i]` 以及周长 `c[i]`。
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接着对威胁最大的病毒区域进行隔离，累加所需的防火墙数量。
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剩余的病毒区域向外感染一个区域。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def containVirus(self, isInfected: List[List[int]]) -> int:
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def dfs(i, j):
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vis[i][j] = True
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areas[-1].add((i, j))
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for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
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x, y = i + a, j + b
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if 0 <= x < m and 0 <= y < n:
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if isInfected[x][y] == 1 and not vis[x][y]:
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dfs(x, y)
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elif isInfected[x][y] == 0:
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c[-1] += 1
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boundaries[-1].add((x, y))
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m, n = len(isInfected), len(isInfected[0])
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ans = 0
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while 1:
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vis = [[False] * n for _ in range(m)]
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areas = []
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c = []
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boundaries = []
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for i, row in enumerate(isInfected):
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for j, v in enumerate(row):
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if v == 1 and not vis[i][j]:
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areas.append(set())
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boundaries.append(set())
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c.append(0)
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dfs(i, j)
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if not areas:
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break
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idx = boundaries.index(max(boundaries, key=len))
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ans += c[idx]
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for k, area in enumerate(areas):
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if k == idx:
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for i, j in area:
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isInfected[i][j] = -1
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else:
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for i, j in area:
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for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
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x, y = i + a, j + b
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if 0 <= x < m and 0 <= y < n and isInfected[x][y] == 0:
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isInfected[x][y] = 1
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private static final int[] DIRS = {-1, 0, 1, 0, -1};
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private List<Integer> c = new ArrayList<>();
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private List<Set<Integer>> areas = new ArrayList<>();
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private List<Set<Integer>> boundaries = new ArrayList<>();
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private int[][] isInfected;
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private int m;
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private int n;
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public int containVirus(int[][] isInfected) {
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m = isInfected.length;
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n = isInfected[0].length;
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this.isInfected = isInfected;
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int ans = 0;
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while (true) {
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boolean[][] vis = new boolean[m][n];
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areas.clear();
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c.clear();
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boundaries.clear();
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (isInfected[i][j] == 1 && !vis[i][j]) {
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areas.add(new HashSet<>());
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boundaries.add(new HashSet<>());
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c.add(0);
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dfs(i, j, vis);
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}
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}
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}
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if (boundaries.isEmpty()) {
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break;
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}
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int idx = max(boundaries);
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ans += c.get(idx);
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for (int t = 0; t < areas.size(); ++t) {
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if (t == idx) {
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for (int v : areas.get(t)) {
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int i = v / n, j = v % n;
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isInfected[i][j] = -1;
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}
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} else {
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for (int v : areas.get(t)) {
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int i = v / n, j = v % n;
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for (int k = 0; k < 4; ++k) {
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int x = i + DIRS[k], y = j + DIRS[k + 1];
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if (x >= 0 && x < m && y >= 0 && y < n && isInfected[x][y] == 0) {
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isInfected[x][y] = 1;
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}
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}
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}
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}
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}
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}
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return ans;
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}
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private int max(List<Set<Integer>> boundaries) {
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int idx = 0;
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int mx = boundaries.get(0).size();
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for (int i = 1; i < boundaries.size(); ++i) {
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int t = boundaries.get(i).size();
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if (mx < t) {
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mx = t;
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idx = i;
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}
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}
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return idx;
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}
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private void dfs(int i, int j, boolean[][] vis) {
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vis[i][j] = true;
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int idx = areas.size() - 1;
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areas.get(idx).add(i * n + j);
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for (int k = 0; k < 4; ++k) {
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int x = i + DIRS[k], y = j + DIRS[k + 1];
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if (x >= 0 && x < m && y >= 0 && y < n) {
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if (isInfected[x][y] == 1 && !vis[x][y]) {
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dfs(x, y, vis);
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} else if (isInfected[x][y] == 0) {
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c.set(idx, c.get(idx) + 1);
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boundaries.get(idx).add(x * n + y);
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}
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}
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}
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}
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}
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```
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### **Go**
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```go
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func containVirus(isInfected [][]int) int {
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m, n := len(isInfected), len(isInfected[0])
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ans := 0
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dirs := []int{-1, 0, 1, 0, -1}
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max := func(boundaries []map[int]bool) int {
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idx := 0
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mx := len(boundaries[0])
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for i, v := range boundaries {
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t := len(v)
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if mx < t {
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mx = t
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idx = i
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}
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}
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return idx
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}
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for {
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vis := make([][]bool, m)
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for i := range vis {
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vis[i] = make([]bool, n)
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}
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areas := []map[int]bool{}
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boundaries := []map[int]bool{}
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c := []int{}
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var dfs func(i, j int)
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dfs = func(i, j int) {
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vis[i][j] = true
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idx := len(areas) - 1
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areas[idx][i*n+j] = true
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for k := 0; k < 4; k++ {
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x, y := i+dirs[k], j+dirs[k+1]
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if x >= 0 && x < m && y >= 0 && y < n {
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if isInfected[x][y] == 1 && !vis[x][y] {
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dfs(x, y)
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} else if isInfected[x][y] == 0 {
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c[idx]++
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boundaries[idx][x*n+y] = true
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}
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}
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}
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}
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for i, row := range isInfected {
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for j, v := range row {
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if v == 1 && !vis[i][j] {
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areas = append(areas, map[int]bool{})
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boundaries = append(boundaries, map[int]bool{})
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c = append(c, 0)
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dfs(i, j)
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}
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}
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}
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if len(areas) == 0 {
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break
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}
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idx := max(boundaries)
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ans += c[idx]
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for t, area := range areas {
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if t == idx {
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for v := range area {
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i, j := v/n, v%n
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isInfected[i][j] = -1
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}
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} else {
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for v := range area {
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i, j := v/n, v%n
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for k := 0; k < 4; k++ {
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x, y := i+dirs[k], j+dirs[k+1]
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if x >= 0 && x < m && y >= 0 && y < n && isInfected[x][y] == 0 {
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isInfected[x][y] = 1
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}
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}
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}
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}
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}
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}
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return ans
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}
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```
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### **...**

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