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type UnionFind struct {
parent []int
// size[i] 表示以 i 为根的树的子节点数目
size []int
}
func NewUnionFind(n int) *UnionFind {
uf := &UnionFind{parent: make([]int, n), size: make([]int, n)}
for i := 0; i < n; i++ {
uf.parent[i] = i
uf.size[i] = 1
}
return uf
}
func (uf *UnionFind) Find(i int) int {
if uf.parent[i] != i {
uf.parent[i] = uf.Find(uf.parent[i])
}
return uf.parent[i]
}
func (uf *UnionFind) Merge(i, j int) {
iF, jF := uf.Find(i), uf.Find(j)
if iF != jF {
uf.parent[iF] = jF
uf.size[jF] += uf.size[iF]
}
}
func (uf *UnionFind) GetSize(i int) int {
return uf.size[uf.Find(i)]
}
func hitBricks(grid [][]int, hits [][]int) []int {
h, w := len(grid), len(grid[0])
replica := make([][]int, h)
for i := 0; i < h; i++ {
replica[i] = make([]int, w)
copy(replica[i], grid[i])
}
for _, hit := range hits {
i, j := hit[0], hit[1]
replica[i][j] = 0
}
getKey := func(i, j int) int {
return i*w + j
}
size := h * w
uf := NewUnionFind(size + 1)
for j := 0; j < w; j++ {
if replica[0][j] == 1 {
uf.Merge(size, j)
}
}
for i := 1; i < h; i++ {
// 不能将 j < w 和 replica[i][j] == 1 这两个条件写在一起，
// 这样写的后果是如果当忽略两当前行后面的1
// for j := 0; j < w && replica[i][j] == 1; j++ {
for j := 0; j < w; j++ {
if replica[i][j] == 1 {
key := getKey(i, j)
if replica[i-1][j] == 1 {
uf.Merge(key, getKey(i-1, j))
}
if j >= 1 && replica[i][j-1] == 1 {
uf.Merge(key, getKey(i, j-1))
}
}
}
}
dirs := [][2]int{{1, 0}, {0, 1}, {-1, 0}, {0, -1}}
ans := make([]int, len(hits))
for i := len(hits) - 1; i >= 0; i-- {
x, y := hits[i][0], hits[i][1]
if grid[x][y] == 0 {
continue
}
originSize := uf.GetSize(size)
if x == 0 {
// 当前要补回的砖块和屋顶相连
uf.Merge(y, size)
}
key := getKey(x, y)
for _, dir := range dirs {
// 向四个方向尝试合并
newX, newY := x+dir[0], y+dir[1]
if 0 <= newX && newX < h && 0 <= newY && newY < w && replica[newX][newY] == 1 {
uf.Merge(key, getKey(newX, newY))
}
}
curSize := uf.GetSize(size)
ans[i] = curSize - originSize - 1
if ans[i] < 0 {
ans[i] = 0
}
// 不要忘记设置 replica[x][y] = 1
replica[x][y] = 1
}
return ans
}
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