[LeetCode] 749. Contain Virus 包含病毒

A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls.

The world is modeled as a 2-D array of cells, where 0 represents uninfected cells, and 1 represents cells contaminated with the virus. A wall (and only one wall) can be installed between any two 4-directionally adjacent cells, on the shared boundary.

Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. Resources are limited. Each day, you can install walls around only one region -- the affected area (continuous block of infected cells) that threatens the most uninfected cells the following night. There will never be a tie.

Can you save the day? If so, what is the number of walls required? If not, and the world becomes fully infected, return the number of walls used.

Example 1:

Input: grid =
[[0,1,0,0,0,0,0,1],
 [0,1,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,1],
 [0,0,0,0,0,0,0,0]]
Output: 10
Explanation:
There are 2 contaminated regions.
On the first day, add 5 walls to quarantine the viral region on the left. The board after the virus spreads is:

[[0,1,0,0,0,0,1,1],
 [0,1,0,0,0,0,1,1],
 [0,0,0,0,0,0,1,1],
 [0,0,0,0,0,0,0,1]]

On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained.

Example 2:

Input: grid =
[[1,1,1],
 [1,0,1],
 [1,1,1]]
Output: 4
Explanation: Even though there is only one cell saved, there are 4 walls built.
Notice that walls are only built on the shared boundary of two different cells.

Example 3:

Input: grid =
[[1,1,1,0,0,0,0,0,0],
 [1,0,1,0,1,1,1,1,1],
 [1,1,1,0,0,0,0,0,0]]
Output: 13
Explanation: The region on the left only builds two new walls.

Note:

1. The number of rows and columns of grid will each be in the range [1, 50].
2. Each grid[i][j] will be either 0 or 1.
3. Throughout the described process, there is always a contiguous viral region that will infect strictly more uncontaminated squares in the next round.

Python:

class Solution(object):
    def containVirus(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]

        def dfs(grid, r, c, lookup, regions, frontiers, perimeters):
            if (r, c) in lookup:
                return
            lookup.add((r, c))
            regions[-1].add((r, c))
            for d in directions:
                nr, nc = r+d[0], c+d[1]
                if not (0 <= nr < len(grid) and \
                        0 <= nc < len(grid[r])):
                    continue
                if grid[nr][nc] == 1:
                    dfs(grid, nr, nc, lookup, regions, frontiers, perimeters)
                elif grid[nr][nc] == 0:
                    frontiers[-1].add((nr, nc))
                    perimeters[-1] += 1

        result = 0
        while True:
            lookup, regions, frontiers, perimeters = set(), [], [], []
            for r, row in enumerate(grid):
                for c, val in enumerate(row):
                    if val == 1 and (r, c) not in lookup:
                        regions.append(set())
                        frontiers.append(set())
                        perimeters.append(0)
                        dfs(grid, r, c, lookup, regions, frontiers, perimeters)

            if not regions: break

            triage_idx = frontiers.index(max(frontiers, key = len))
            for i, region in enumerate(regions):
                if i == triage_idx:
                    result += perimeters[i]
                    for r, c in region:
                        grid[r][c] = -1
                    continue
                for r, c in region:
                    for d in directions:
                        nr, nc = r+d[0], c+d[1]
                        if not (0 <= nr < len(grid) and \
                                0 <= nc < len(grid[r])):
                            continue
                        if grid[nr][nc] == 0:
                            grid[nr][nc] = 1

        return result

C++:

class Solution {
public:
    int containVirus(vector<vector<int>>& grid) {
        int res = 0, m = grid.size(), n = grid[0].size();
        vector<vector<int>> dirs{{-1,0},{0,1},{1,0},{0,-1}};
        while (true) {
            unordered_set<int> visited;
            vector<vector<vector<int>>> all;
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++j) {
                    if (grid[i][j] == 1 && !visited.count(i * n + j)) {
                        queue<int> q{{i * n + j}};
                        vector<int> virus{i * n + j};
                        vector<int> walls;
                        visited.insert(i * n + j);
                        while (!q.empty()) {
                            auto t = q.front(); q.pop();
                            for (auto dir : dirs) {
                                int x = (t / n) + dir[0], y = (t % n) + dir[1];
                                if (x < 0 || x >= m || y < 0 || y >= n || visited.count(x * n + y)) continue;
                                if (grid[x][y] == -1) continue;
                                else if (grid[x][y] == 0) walls.push_back(x * n + y);
                                else if (grid[x][y] == 1) {
                                    visited.insert(x * n + y);
                                    virus.push_back(x * n + y);
                                    q.push(x * n + y);
                                }
                            }
                        }
                        unordered_set<int> s(walls.begin(), walls.end());
                        vector<int> cells{(int)s.size()};
                        all.push_back({cells ,walls, virus});
                    }
                }
            }
            if (all.empty()) break;
            sort(all.begin(), all.end(), [](vector<vector<int>> &a, vector<vector<int>> &b) {return a[0][0] > b[0][0];});
            for (int i = 0; i < all.size(); ++i) {
                if (i == 0) {
                    vector<int> virus = all[0][2];
                    for (int idx : virus) grid[idx / n][idx % n] = -1;
                    res += all[0][1].size();
                } else {
                    vector<int> wall = all[i][1];
                    for (int idx : wall) grid[idx / n][idx % n] = 1;
                }
            }
        }
        return res;
    }
};

　　　

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