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segmentTree.cpp
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218 lines (213 loc) · 7.81 KB
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#include<vector>
#include<iostream>
#include<unordered_map>
#include<set>
using namespace std;
struct Node {//[l,r)
int l;
int r;
int height;
Node() : l(0), r(0), height(0) {}
};
class Tree {
private:
Node* nodes;
int s;
public:
vector<vector<int>> ans;
vector<int> temp;
Tree(int size) {
s = size;
temp.resize(2);
nodes = new Node[4 * size];
nodes[1].l = 0;
nodes[1].r = size - 1;
int father, left, right, mid;
for (int i = 2; i < 4 * size - 1; i += 2) {
father = (i >> 1);
left = nodes[father].l, right = nodes[father].r;
if (left < right - 1) {
mid = (left + right) >> 1;
nodes[i].l = left;
nodes[i].r = mid;
nodes[i + 1].l = mid;
nodes[i + 1].r = right;
}
}
}
~Tree() {
delete[] nodes;
}
void print() {
for (int i = 1; i < 4 * s; i++) {
if (nodes[i].l == nodes[i].r - 1)
cout << nodes[i].l << " " << nodes[i].r << " " << nodes[i].height << endl;
}
}
void update(int l, int r, int height, int index) {
if (nodes[index].l >= r || nodes[index].r <= l)
return;
if (nodes[index].l >= l && nodes[index].r <= r) {
nodes[index].height = nodes[index].height > height ? nodes[index].height : height;
return;
}
if (nodes[index].l == nodes[index].r - 1)
return;
if (nodes[index].height != 0) {
nodes[index << 1].height = nodes[index << 1].height > nodes[index].height ? nodes[index << 1].height : nodes[index].height;
nodes[(index << 1) + 1].height = nodes[(index << 1) + 1].height > nodes[index].height ? nodes[(index << 1) + 1].height : nodes[index].height;
nodes[index].height = 0;
}
update(l, r, height, index << 1);
update(l, r, height, (index << 1) + 1);
}
void clear() {
for (int index = 1; index < 4 * s; index++) {
if (nodes[index].l < nodes[index].r - 1 && nodes[index].height) {
nodes[index << 1].height = nodes[index << 1].height > nodes[index].height ? nodes[index << 1].height : nodes[index].height;
nodes[(index << 1) + 1].height = nodes[(index << 1) + 1].height > nodes[index].height ? nodes[(index << 1) + 1].height : nodes[index].height;
nodes[index].height = 0;
}
}
}
vector<vector<int>> getAns() {
sum(1);
return ans;
}
void sum(int index) {
if (nodes[index].l == nodes[index].r - 1) {
if(ans[ans.size() - 1][1] != nodes[index].height) {
temp[0] = nodes[index].l;
temp[1] = nodes[index].height;
ans.push_back(temp);
}
}
else {
sum(2*index);
sum(2*index + 1);
}
}
};
class Solution {
public:
vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
if (buildings.size() == 0)
return {};
unordered_map<int,int> locToIndex;
vector<int> indexToLoc;
int num = 0;
set<int> locs;
for (auto x : buildings) {
locs.insert(x[0]);
locs.insert(x[1]);
}
for (int x : locs) {
indexToLoc.push_back(x);
locToIndex[x] = num++;
}
Tree tree(num);
for (auto x : buildings) {
tree.update(locToIndex[x[0]], locToIndex[x[1]], x[2], 1);
}
tree.clear();
tree.print();
vector<vector<int>> ans = tree.getAns();
// for (int i = 0; i < ans.size() ;i++) {
// ans[i][0] = indexToLoc[ans[i][0]];
// }
return ans;
}
};
//注意在修改过程中学会根据修改的形式进行随机应变,如,如果只进行单个元素的修改,b数组根本就没有什么用处。
//本实现中以1为根节点,从始至终为闭集进行操作。p为当前节点下标,s,t在建树时为原数组,在查询和更新时为树的节点。
//当心更新值为0的问题。
class segmentTree
{
private:
int* d;
int* b;
public:
segmentTree(vector<int> a) {
d = new int[4 * a.size()];
b = new int[4 * a.size()];
for (int i = 0; i < 4 * a.size(); i++) {
b[i] = 0;
}
build(0, a.size() - 1, 1, a);
}
void build(int s, int t, int p, vector<int> a) {
// 对 [s,t] 区间建立线段树,当前根的编号为 p
if (s == t) {
d[p] = a[s];
return;
}
int m = s + ((t - s) >> 1);
// 移位运算符的优先级小于加减法,所以加上括号
// 如果写成 (s + t) >> 1 可能会时间超限
build(s, m, p * 2, a), build(m + 1, t, p * 2 + 1, a);
// 递归对左右区间建树
d[p] = d[p * 2] + d[(p * 2) + 1];
}
// void update(int l, int r, int c, int s, int t, int p) {
// // [l,r] 为修改区间,c 为被修改的元素的变化量,[s,t] 为当前节点包含的区间,p
// // 为当前节点的编号
// if (l <= s && t <= r) {
// d[p] += (t - s + 1) * c, b[p] += c;
// return;
// } // 当前区间为修改区间的子集时直接修改当前节点的值,然后打标记,结束修改
// int m = s + ((t - s) >> 1);
// if (b[p] && s != t) {
// // 如果当前节点的懒标记非空,则更新当前节点两个子节点的值和懒标记值
// d[p * 2] += b[p] * (m - s + 1), d[p * 2 + 1] += b[p] * (t - m);
// b[p * 2] += b[p], b[p * 2 + 1] += b[p]; // 将标记下传给子节点
// b[p] = 0; // 清空当前节点的标记
// }
// if (l <= m) update(l, r, c, s, m, p * 2);
// if (r > m) update(l, r, c, m + 1, t, p * 2 + 1);
// d[p] = d[p * 2] + d[p * 2 + 1];
// }
// int getsum(int l, int r, int s, int t, int p) {
// // [l,r] 为查询区间,[s,t] 为当前节点包含的区间,p为当前节点的编号
// if (l <= s && t <= r) return d[p];
// // 当前区间为询问区间的子集时直接返回当前区间的和
// int m = s + ((t - s) >> 1);
// if (b[p]) {
// // 如果当前节点的懒标记非空,则更新当前节点两个子节点的值和懒标记值
// d[p * 2] += b[p] * (m - s + 1), d[p * 2 + 1] += b[p] * (t - m),
// b[p * 2] += b[p], b[p * 2 + 1] += b[p]; // 将标记下传给子节点
// b[p] = 0; // 清空当前节点的标记
// }
// int sum = 0;
// if (l <= m) sum = getsum(l, r, s, m, p * 2);
// if (r > m) sum += getsum(l, r, m + 1, t, p * 2 + 1);
// return sum;
// }
void update(int l, int r, int c, int s, int t, int p) {
if (l <= s && t <= r) {
d[p] = (t - s + 1) * c, b[p] = c;
return;
}
int m = s + ((t - s) >> 1);
if (b[p]) {
d[p * 2] = b[p] * (m - s + 1), d[p * 2 + 1] = b[p] * (t - m),
b[p * 2] = b[p * 2 + 1] = b[p];
b[p] = 0;
}
if (l <= m) update(l, r, c, s, m, p * 2);
if (r > m) update(l, r, c, m + 1, t, p * 2 + 1);
d[p] = d[p * 2] + d[p * 2 + 1];
}
int getsum(int l, int r, int s, int t, int p) {
if (l <= s && t <= r) return d[p];
int m = s + ((t - s) >> 1);
if (b[p]) {
d[p * 2] = b[p] * (m - s + 1), d[p * 2 + 1] = b[p] * (t - m),
b[p * 2] = b[p * 2 + 1] = b[p];
b[p] = 0;
}
int sum = 0;
if (l <= m) sum = getsum(l, r, s, m, p * 2);
if (r > m) sum += getsum(l, r, m + 1, t, p * 2 + 1);
return sum;
}
};