Leetcode—SkyLine Problem

问题描述：

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], ...]

问题解读：

　　这个问题理解下来就是给了一堆的矩形，要你求轮廓线。然而弱渣的我一开始只能想到x是整数所以可以暴力的遍历来做这道题，然而想都不用想肯定会跪，想了一会没有十分好的思路之后，看了一看大神们的博客，终于有了思路，决定使用优先队列。在这里使用的优先队列里面存的是一组int型的pair，第一个存矩形的高度，第二个存矩形的x2坐标。优先队列的特性是会按照pair的第一个的大小进行排序。

算法描述：

　　首先定义一个空的优先队列和一个存最后结果的vector。n定义为输入向量的长度。输入向量a，里面存的向量记为（x0,x1,h)

　　然后循环，i=0，循环条件i<n或者优先队列不为空。

　　当队列为空或者此时的x0小于队列顶端的x1时就更新x，添加队列元素

　　否则更新x，pop队列元素，

　　之后再根据情况更新x，y（语文太捉急了还是直接上代码吧）

class Solution {
public:
  vector<pair<int, int>> getSkyline(vector<vector<int>>& a) {
    int n = a.size(), xx = -1, x, yy = 0, y = 0;
    priority_queue<pair<int, int>> pq; //根据第一个元素排序
    vector<pair<int, int>> r;//用于储存结果
    for (int i = 0; i < n || pq.size(); ) {
      //如果新矩形的左端点在一个更大的矩形内部，或者上面没有矩形覆盖
      if (pq.empty() || i < n && a[i][0] < pq.top().second) {
        x = a[i][0];
        pq.emplace(a[i][2], a[i][1]);
        i++;
      } else {
          //如果新矩形是上没有矩形覆盖
        x = pq.top().second;
        while (pq.size() && pq.top().second <= x)
          pq.pop();
      }
      //判断是否有更新端点
      if (x != xx && y != yy) {
        r.emplace_back(xx, y);
        yy = y;
      }
      xx = x;
      y = pq.empty() ? 0 : pq.top().first;
    }
    //防止最后出现0点
    if (y != yy)
      r.emplace_back(xx, y);
    return r;
  }
};