LWC 63: 749. Contain Virus
LWC 63: 749. Contain Virus
Problem:
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:
The number of rows and columns of grid will each be in the range [1, 50].
Each grid[i][j] will be either 0 or 1.
Throughout the described process, there is always a contiguous viral region that will infect strictly more uncontaminated squares in the next round.
思路： 此题实际上是一个模拟＋贪心，首先可以简单计算每次感染后得到的边界大小。优先选择感染边界最大的，这样可以减少后续的感染人数。接下来，只需要确定每次感染后的图像，根据这些图像重新一轮计算，直到没有新的区域产生。
Java版本：
public class Solution {
Set<Integer> vis;
List<Set<Integer>> frontiers;
List<Set<Integer>> regions;
List<Integer> perimeters;
int[][] grid;
int n, m;
int[][] dir = {{0, 1},{1, 0},{0, -1},{-1, 0}};
public int containVirus(int[][] grid) {
this.grid = grid;
n = grid.length;
m = grid[0].length;
int ans = 0;
while (true) {
vis = new HashSet<>();
frontiers = new ArrayList<>();
regions = new ArrayList<>();
perimeters = new ArrayList<>();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] ==1 && !vis.contains(i * m + j)) {
frontiers.add(new HashSet<>());
regions.add(new HashSet<>());
perimeters.add(0);
dfs(i, j);
}
}
}
if (regions.isEmpty()) break;
int triageIndex = 0;
for (int i = 0; i < frontiers.size(); ++i) {
if (frontiers.get(triageIndex).size() < frontiers.get(i).size())
triageIndex = i;
}
ans += perimeters.get(triageIndex);
for (int i = 0; i < regions.size(); ++i) {
if (i == triageIndex) {
for (int code: regions.get(i))
grid[code / m][code % m] = -1;
} else {
for (int code: regions.get(i)) {
int r = code / m, c = code % m;
for (int[] d : dir) {
int nr = r + d[0], nc = c + d[1];
if (nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] == 0)
grid[nr][nc] = 1;
}
}
}
}
}
return ans;
}
public void dfs(int r, int c) {
if (!vis.contains(r * m + c)) {
vis.add(r * m + c);
int N = regions.size();
regions.get(N - 1).add(r * m + c);
for (int[] d : dir) {
int nr = d[0] + r;
int nc = d[1] + c;
if (nr >= 0 && nr < n && nc >= 0 && nc < m) {
if (grid[nr][nc] == 1) {
dfs(nr, nc);
}
else if (grid[nr][nc] == 0){
frontiers.get(N - 1).add(nr * m + nc);
perimeters.set(N - 1, perimeters.get(N - 1) + 1);
}
}
}
}
}
}Python版本：
class Solution(object):
def containVirus(self, grid):
R, C = len(grid), len(grid[0])
def neighbors(r, c):
for nr, nc in ((r-1, c), (r+1, c), (r, c-1), (r, c+1)):
if 0 <= nr < R and 0 <= nc < C:
yield nr, nc
def dfs(r, c):
if (r, c) not in seen:
seen.add((r, c))
regions[-1].add((r, c))
for nr, nc in neighbors(r, c):
if grid[nr][nc] == 1:
dfs(nr, nc)
elif grid[nr][nc] == 0:
frontiers[-1].add((nr, nc))
perimeters[-1] += 1
ans = 0
while True:
#Find all regions, with associated frontiers and perimeters.
seen = set()
regions = []
frontiers = []
perimeters = []
for r, row in enumerate(grid):
for c, val in enumerate(row):
if val == 1 and (r, c) not in seen:
regions.append(set())
frontiers.append(set())
perimeters.append(0)
dfs(r, c)
if not regions: break
triage_index = frontiers.index(max(frontiers, key = len))
ans += perimeters[triage_index]
for i, reg in enumerate(regions):
if i == triage_index:
for r, c in reg:
grid[r][c] = -1
else:
for r, c in reg:
for nr, nc in neighbors(r, c):
if grid[nr][nc] == 0:
grid[nr][nc] = 1
return ans