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困难 |
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第 285 场周赛 Q4 |
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给你一个下标从 0 开始的字符串 s 。另给你一个下标从 0 开始、长度为 k 的字符串 queryCharacters ，一个下标从 0 开始、长度也是 k 的整数 下标 数组 queryIndices ，这两个都用来描述 k 个查询。
第 i 个查询会将 s 中位于下标 queryIndices[i] 的字符更新为 queryCharacters[i] 。
返回一个长度为 k 的数组 lengths ，其中 lengths[i] 是在执行第 i 个查询 之后 s 中仅由 单个字符重复 组成的 最长子字符串 的 长度 。
示例 1：
输入：s = "babacc", queryCharacters = "bcb", queryIndices = [1,3,3] 输出：[3,3,4] 解释： - 第 1 次查询更新后 s = "bbbacc" 。由单个字符重复组成的最长子字符串是 "bbb" ，长度为 3 。 - 第 2 次查询更新后 s = "bbbccc" 。由单个字符重复组成的最长子字符串是 "bbb" 或 "ccc"，长度为 3 。 - 第 3 次查询更新后 s = "bbbbcc" 。由单个字符重复组成的最长子字符串是 "bbbb" ，长度为 4 。 因此，返回 [3,3,4] 。
示例 2：
输入：s = "abyzz", queryCharacters = "aa", queryIndices = [2,1] 输出：[2,3] 解释： - 第 1 次查询更新后 s = "abazz" 。由单个字符重复组成的最长子字符串是 "zz" ，长度为 2 。 - 第 2 次查询更新后 s = "aaazz" 。由单个字符重复组成的最长子字符串是 "aaa" ，长度为 3 。 因此，返回 [2,3] 。
提示：
1 <= s.length <= 105s由小写英文字母组成k == queryCharacters.length == queryIndices.length1 <= k <= 105queryCharacters由小写英文字母组成0 <= queryIndices[i] < s.length
线段树将整个区间分割为多个不连续的子区间，子区间的数量不超过 log(width)。更新某个元素的值，只需要更新 log(width) 个区间，并且这些区间都包含在一个包含该元素的大区间内。区间修改时，需要使用懒标记保证效率。
- 线段树的每个节点代表一个区间；
- 线段树具有唯一的根节点，代表的区间是整个统计范围，如
[1, N]； - 线段树的每个叶子节点代表一个长度为 1 的元区间
[x, x]； - 对于每个内部节点
[l, r]，它的左儿子是[l, mid]，右儿子是[mid + 1, r], 其中mid = ⌊(l + r) / 2⌋(即向下取整)。
对于本题，线段树节点维护的信息有：
- 前缀最长连续字符个数
lmx； - 后缀最长连续字符个数
rmx； - 区间最长连续字符个数
mx； - 区间长度
size； - 区间首尾字符
lc, rc。
class Node:
def __init__(self):
self.l = 0
self.r = 0
self.lmx = 0
self.rmx = 0
self.mx = 0
self.size = 0
self.lc = None
self.rc = None
N = 100010
tr = [Node() for _ in range(N << 2)]
class SegmentTree:
def __init__(self, s):
n = len(s)
self.s = s
self.build(1, 1, n)
def build(self, u, l, r):
tr[u].l = l
tr[u].r = r
if l == r:
tr[u].lmx = tr[u].rmx = tr[u].mx = tr[u].size = 1
tr[u].lc = tr[u].rc = self.s[l - 1]
return
mid = (l + r) >> 1
self.build(u << 1, l, mid)
self.build(u << 1 | 1, mid + 1, r)
self.pushup(u)
def modify(self, u, x, v):
if tr[u].l == x and tr[u].r == x:
tr[u].lc = tr[u].rc = v
return
mid = (tr[u].l + tr[u].r) >> 1
if x <= mid:
self.modify(u << 1, x, v)
else:
self.modify(u << 1 | 1, x, v)
self.pushup(u)
def query(self, u, l, r):
if tr[u].l >= l and tr[u].r <= r:
return tr[u]
mid = (tr[u].l + tr[u].r) >> 1
if r <= mid:
return self.query(u << 1, l, r)
if l > mid:
return self.query(u << 1 | 1, l, r)
left, right = self.query(u << 1, l, r), self.query(u << 1 | 1, l, r)
ans = Node()
self._pushup(ans, left, right)
return ans
def _pushup(self, root, left, right):
root.lc, root.rc = left.lc, right.rc
root.size = left.size + right.size
root.mx = max(left.mx, right.mx)
root.lmx, root.rmx = left.lmx, right.rmx
if left.rc == right.lc:
if left.lmx == left.size:
root.lmx += right.lmx
if right.rmx == right.size:
root.rmx += left.rmx
root.mx = max(root.mx, left.rmx + right.lmx)
def pushup(self, u):
self._pushup(tr[u], tr[u << 1], tr[u << 1 | 1])
class Solution:
def longestRepeating(
self, s: str, queryCharacters: str, queryIndices: List[int]
) -> List[int]:
tree = SegmentTree(s)
k = len(queryIndices)
ans = []
for i, c in enumerate(queryCharacters):
x = queryIndices[i] + 1
tree.modify(1, x, c)
ans.append(tree.query(1, 1, len(s)).mx)
return ansclass Node {
int l;
int r;
int size;
int lmx;
int rmx;
int mx;
char lc;
char rc;
}
class SegmentTree {
private String s;
private Node[] tr;
public SegmentTree(String s) {
int n = s.length();
this.s = s;
tr = new Node[n << 2];
for (int i = 0; i < tr.length; ++i) {
tr[i] = new Node();
}
build(1, 1, n);
}
public void build(int u, int l, int r) {
tr[u].l = l;
tr[u].r = r;
if (l == r) {
tr[u].lmx = 1;
tr[u].rmx = 1;
tr[u].mx = 1;
tr[u].size = 1;
tr[u].lc = s.charAt(l - 1);
tr[u].rc = s.charAt(l - 1);
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
pushup(u);
}
void modify(int u, int x, char v) {
if (tr[u].l == x && tr[u].r == x) {
tr[u].lc = v;
tr[u].rc = v;
return;
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (x <= mid) {
modify(u << 1, x, v);
} else {
modify(u << 1 | 1, x, v);
}
pushup(u);
}
Node query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) {
return tr[u];
}
int mid = (tr[u].l + tr[u].r) >> 1;
if (r <= mid) {
return query(u << 1, l, r);
}
if (l > mid) {
return query(u << 1 | 1, l, r);
}
Node ans = new Node();
Node left = query(u << 1, l, r);
Node right = query(u << 1 | 1, l, r);
pushup(ans, left, right);
return ans;
}
void pushup(Node root, Node left, Node right) {
root.lc = left.lc;
root.rc = right.rc;
root.size = left.size + right.size;
root.mx = Math.max(left.mx, right.mx);
root.lmx = left.lmx;
root.rmx = right.rmx;
if (left.rc == right.lc) {
if (left.lmx == left.size) {
root.lmx += right.lmx;
}
if (right.rmx == right.size) {
root.rmx += left.rmx;
}
root.mx = Math.max(root.mx, left.rmx + right.lmx);
}
}
void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
}
class Solution {
public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {
SegmentTree tree = new SegmentTree(s);
int k = queryCharacters.length();
int[] ans = new int[k];
for (int i = 0; i < k; ++i) {
int x = queryIndices[i] + 1;
char c = queryCharacters.charAt(i);
tree.modify(1, x, c);
ans[i] = tree.query(1, 1, s.length()).mx;
}
return ans;
}
}class Node {
public:
int l, r, size, lmx, rmx, mx;
char lc, rc;
};
class SegmentTree {
private:
string s;
vector<Node*> tr;
public:
SegmentTree(string& s) {
this->s = s;
int n = s.size();
tr.resize(n << 2);
for (int i = 0; i < tr.size(); ++i) tr[i] = new Node();
build(1, 1, n);
}
void build(int u, int l, int r) {
tr[u]->l = l;
tr[u]->r = r;
if (l == r) {
tr[u]->lmx = tr[u]->rmx = tr[u]->mx = tr[u]->size = 1;
tr[u]->lc = tr[u]->rc = s[l - 1];
return;
}
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
pushup(u);
}
void modify(int u, int x, char v) {
if (tr[u]->l == x && tr[u]->r == x) {
tr[u]->lc = tr[u]->rc = v;
return;
}
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (x <= mid)
modify(u << 1, x, v);
else
modify(u << 1 | 1, x, v);
pushup(u);
}
Node* query(int u, int l, int r) {
if (tr[u]->l >= l && tr[u]->r <= r) return tr[u];
int mid = (tr[u]->l + tr[u]->r) >> 1;
if (r <= mid) return query(u << 1, l, r);
if (l > mid) query(u << 1 | 1, l, r);
Node* ans = new Node();
Node* left = query(u << 1, l, r);
Node* right = query(u << 1 | 1, l, r);
pushup(ans, left, right);
return ans;
}
void pushup(Node* root, Node* left, Node* right) {
root->lc = left->lc;
root->rc = right->rc;
root->size = left->size + right->size;
root->mx = max(left->mx, right->mx);
root->lmx = left->lmx;
root->rmx = right->rmx;
if (left->rc == right->lc) {
if (left->lmx == left->size) root->lmx += right->lmx;
if (right->rmx == right->size) root->rmx += left->rmx;
root->mx = max(root->mx, left->rmx + right->lmx);
}
}
void pushup(int u) {
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
};
class Solution {
public:
vector<int> longestRepeating(string s, string queryCharacters, vector<int>& queryIndices) {
SegmentTree* tree = new SegmentTree(s);
int k = queryCharacters.size();
vector<int> ans(k);
for (int i = 0; i < k; ++i) {
int x = queryIndices[i] + 1;
tree->modify(1, x, queryCharacters[i]);
ans[i] = tree->query(1, 1, s.size())->mx;
}
return ans;
}
};type segmentTree struct {
str []byte
mx []int
lmx []int
rmx []int
}
func newSegmentTree(s string) *segmentTree {
n := len(s)
t := &segmentTree{
str: []byte(s),
mx: make([]int, n<<2),
lmx: make([]int, n<<2),
rmx: make([]int, n<<2),
}
t.build(0, 0, n-1)
return t
}
func (t *segmentTree) build(x, l, r int) {
if l == r {
t.lmx[x] = 1
t.rmx[x] = 1
t.mx[x] = 1
return
}
m := int(uint(l+r) >> 1)
t.build(x*2+1, l, m)
t.build(x*2+2, m+1, r)
t.pushup(x, l, m, r)
}
func (t *segmentTree) pushup(x, l, m, r int) {
lch, rch := x*2+1, x*2+2
t.lmx[x] = t.lmx[lch]
t.rmx[x] = t.rmx[rch]
t.mx[x] = max(t.mx[lch], t.mx[rch])
// can be merged
if t.str[m] == t.str[m+1] {
if t.lmx[lch] == m-l+1 {
t.lmx[x] += t.lmx[rch]
}
if t.rmx[rch] == r-m {
t.rmx[x] += t.rmx[lch]
}
t.mx[x] = max(t.mx[x], t.rmx[lch]+t.lmx[rch])
}
}
func (t *segmentTree) update(x, l, r, pos int, val byte) {
if l == r {
t.str[pos] = val
return
}
m := int(uint(l+r) >> 1)
if pos <= m {
t.update(x*2+1, l, m, pos, val)
} else {
t.update(x*2+2, m+1, r, pos, val)
}
t.pushup(x, l, m, r)
}
func longestRepeating(s string, queryCharacters string, queryIndices []int) []int {
ans := make([]int, len(queryCharacters))
t := newSegmentTree(s)
n := len(s)
for i, c := range queryCharacters {
t.update(0, 0, n-1, queryIndices[i], byte(c))
ans[i] = t.mx[0]
}
return ans
}