# Skyline Problems

14 Feb 2016

A city’s skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you are given the locations and height of all the buildings as shown on a cityscape photo (Figure A), write a program to output the skyline formed by these buildings collectively (Figure B).

The geometric information of each building is represented by a triplet of integers [Li, Ri, Hi], where Li and Ri are the x coordinates of the left and right edge of the ith building, respectively, and Hi is its height. It is guaranteed that 0 ≤ Li, Ri ≤ INT_MAX, 0 < Hi ≤ INT_MAX, and Ri - Li > 0.

For instance, the dimensions of all buildings in Figure A are recorded as: [ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ]

The output is a list of “key points” (red dots in Figure B) in the format of [ [x1,y1], [x2, y2], [x3, y3], ... ] that uniquely defines a skyline. A key point is the left endpoint of a horizontal line segment.

For instance, the skyline in Figure B should be represented as: [ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ].

More details on Leetcode 218 The skyline problem.

Idea:

• sort the start and end points for all the buildings and check each point one by one.
• use a heap to store the heights of current buildings, i.e. overlapped buildings.
• use pre and cur to store the heights of previous building and current building respectively. if pre!=cur output the result.

Code:

Another similar LintCode problem Building Outline

• the only difference is how to output the result.

Code:

vector<vector<int> > buildingOutline(vector<vector<int> > &buildings) {
vector<pair<int, int>> height;
for (auto b : buildings) {
height.push_back(make_pair(b[0], -b[2]));
height.push_back(make_pair(b[1],  b[2]));
}

sort(height.begin(), height.end());
multiset<int> heap;
heap.insert(0);
vector<vector<int>> res;
int pre = 0, cur = 0, start = 0;

for (auto h : height) {
if (h.second < 0) {
heap.insert(-h.second);
} else {
heap.erase(heap.find(h.second));
}
cur = *heap.rbegin();
if (cur != pre) {
if (pre != 0)
res.push_back({start, h.first, pre});
start = h.first;
}
pre = cur;
}
return res;
}

Enjoy the problems.