leetcode 218. The Skyline Problem

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. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 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. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.

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] ].

Notes:

• The number of buildings in any input list is guaranteed to be in the range [0, 10000].
• The input list is already sorted in ascending order by the left x position Li. 
• The output list must be sorted by the x position. 
• There must be no consecutive horizontal lines of equal height in the output skyline. For instance, [...[2 3], [4 5], [7 5], [11 5], [12 7]...] is not acceptable; the three lines of height 5 should be merged into one in the final output as such: [...[2 3], [4 5], [12 7], ...]

分析：

使用扫描线的方法，将一个建筑分为进入和退出，priority_queue<t_point> tt来存这些事件，可以按照time来排序，

flag为1代表进入，flag为2代表退出。分别讨论。

map<int, int, greater<int>> hh作为大顶堆，来存还未退出的建筑的高度，这样就知道当前扫描线上有哪些高度的建筑。

class t_point{public:    int time;    int high;    int flag;        t_point(int a,int b,int d)    {        this->time = a;        this->high = b;        this->flag = d;    }        bool operator<(const t_point b) const    {        if (this->time > b.time)            return true;        else if (this->time < b.time)            return false;        else        {            if (this->flag == 1 && b.flag == 2)                return false;            else if (this->flag == 2 && b.flag == 1)                return true;            else if (this->flag == 1 && b.flag == 1)                return this->high < b.high;            else if (this->flag == 2 && b.flag == 2)                return this->high > b.high;                    }        }};class Solution {public:    vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings)     {        priority_queue<t_point> tt; //可以自动按时间顺序排序        for (int i = 0; i < buildings.size(); i++)        {            tt.push(t_point(buildings[i][0], buildings[i][2], 1));             tt.push(t_point(buildings[i][1], buildings[i][2], 2));        }                map<int, int, greater<int>> hh; //大顶堆，第一个int存还没退出的高度        vector<pair<int, int>> ret;                while (!tt.empty())        {            auto it = tt.top();            tt.pop();                        cout<<it.time<<" "<<it.high<<" "<<it.flag<<endl;                        if (it.flag == 1) //in            {                if (hh.empty() || it.high > hh.begin()->first) //当前没有建筑 或者 进来的建筑比当前最高的建筑还高                {                    ret.push_back(make_pair(it.time, it.high));                }                                   hh[it.high] ++;            }            else //out            {                if (--hh[it.high] == 0)  //hh中退出当前退出的高度                    hh.erase(it.high);                                if (hh.empty()) //如果当前退出之后没有建筑了，这个点肯定要push进ret                {                    ret.push_back(make_pair(it.time, hh.begin()->first));                       continue;                }                                if (it.high > hh.begin()->first) //如果当前建筑退出后，最高的高度(hh.begin()->first)下降了。说明这个点肯定要push进ret                {                    ret.push_back(make_pair(it.time, hh.begin()->first));                }            }        }        return ret;            }};