chapter-four/0800~0899/0850.Rectangle-Area-II

850. Rectangle Area II

题目

We are given a list of (axis-aligned) rectangles. Each rectangle[i] = [x1, y1, x2, y2] , where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner of the ith rectangle.

Find the total area covered by all rectangles in the plane. Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: [[0,0,2,2],[1,0,2,3],[1,0,3,1]]
Output: 6
Explanation: As illustrated in the picture.

Example 2:

Input: [[0,0,1000000000,1000000000]]
Output: 49
Explanation: The answer is 10^18 modulo (10^9 + 7), which is (10^9)^2 = (-7)^2 = 49.

Note:

  • 1 <= rectangles.length <= 200
  • rectanges[i].length = 4
  • 0 <= rectangles[i][j] <= 10^9
  • The total area covered by all rectangles will never exceed 2^63 - 1 and thus will fit in a 64-bit signed integer.

题目大意

我们给出了一个（轴对齐的）矩形列表 rectangles。 对于 rectangle[i] = [x1, y1, x2, y2]，其中（x1，y1）是矩形 i 左下角的坐标，（x2，y2）是该矩形右上角的坐标。找出平面中所有矩形叠加覆盖后的总面积。由于答案可能太大，请返回它对 10 ^ 9 + 7 取模的结果。

提示：

  • 1 <= rectangles.length <= 200
  • rectanges[i].length = 4
  • 0 <= rectangles[i][j] <= 10^9
  • 矩形叠加覆盖后的总面积不会超越 2^63 - 1 ，这意味着可以用一个 64 位有符号整数来保存面积结果。

解题思路

  • 在二维坐标系中给出一些矩形，要求这些矩形合并之后的面积。由于矩形有重叠，所以需要考虑合并以后的面积。矩形的坐标值也会很大。

  • 这一题给人的感觉很像第 218 题，求天际线的过程也是有楼挡楼，重叠的情况。不过那一题只用求天际线的拐点，所以我们可以对区间做“右边界减一”的处理，防止两个相邻区间因为共点，而导致结果错误。但是这一题如果还是用相同的做法，就会出错，因为“右边界减一”以后，面积会少一部分，最终得到的结果也是偏小的。所以这一题要将线段树改造一下。

  • 思路是先讲 Y 轴上的坐标离线化，转换成线段树。将矩形的 2 条边变成扫描线，左边是入边，右边是出边。

  • 再从左往右遍历每条扫描线，并对 Y 轴上的线段树进行 update。X 轴上的每个坐标区间 * query 线段树总高度的结果 = 区间面积。最后将 X 轴对应的每个区间面积加起来，就是最终矩形合并以后的面积。如下图中间的图。

    需要注意的一点是，每次 query 的结果并不一定是连续线段。如上图最右边的图，中间有一段是可能出现镂空的。这种情况看似复杂，其实很简单，因为每段线段树的线段代表的权值高度是不同的，每次 query 最大高度得到的结果已经考虑了中间可能有镂空的情况了。

  • 具体做法，先把各个矩形在 Y 轴方向上离散化，这里的线段树叶子节点不再是一个点了，而是一个区间长度为 1 的区间段。

    每个叶子节点也不再是存储一个 int 值，而是存 2 个值，一个是 count 值，用来记录这条区间被覆盖的次数，另一个值是 val 值，用来反映射该线段长度是多少，因为 Y 轴被离散化了，区间坐标间隔都是 1，但是实际 Y 轴的高度并不是 1 ，所以用 val 来反映射原来的高度。

  • 初始化线段树，叶子节点的 count = 0，val 根据题目给的 Y 坐标进行计算。

  • 从左往右遍历每个扫描线。每条扫面线都把对应 update 更新到叶子节点。pushUp 的时候需要合并每个区间段的高度 val 值。如果有区间没有被覆盖，那么这个区间高度 val 为 0，这也就处理了可能“中间镂空”的情况。

      func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
  	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
  	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
  		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
  	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
  		newValue = sat.tree[leftTreeIndex].val
  	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
  		newValue = sat.tree[rightTreeIndex].val
  	}
  	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
  }

  • 扫描每一个扫描线，先 pushDown 到叶子节点，再 pushUp 到根节点。

  • 遍历到倒数第 2 根扫描线的时候就能得到结果了。因为最后一根扫描线 update 以后，整个线段树全部都归为初始化状态了。

  • 这一题是线段树扫面线解法的经典题。

代码


package leetcode

import (
	"sort"
)

func rectangleArea(rectangles [][]int) int {
	sat, res := SegmentAreaTree{}, 0
	posXMap, posX, posYMap, posY, lines := discretization850(rectangles)
	tmp := make([]int, len(posYMap))
	for i := 0; i < len(tmp)-1; i++ {
		tmp[i] = posY[i+1] - posY[i]
	}
	sat.Init(tmp, func(i, j int) int {
		return i + j
	})
	for i := 0; i < len(posY)-1; i++ {
		tmp[i] = posY[i+1] - posY[i]
	}
	for i := 0; i < len(posX)-1; i++ {
		for _, v := range lines[posXMap[posX[i]]] {
			sat.Update(posYMap[v.start], posYMap[v.end], v.state)
		}
		res += ((posX[i+1] - posX[i]) * sat.Query(0, len(posY)-1)) % 1000000007
	}
	return res % 1000000007
}

func discretization850(positions [][]int) (map[int]int, []int, map[int]int, []int, map[int][]LineItem) {
	tmpXMap, tmpYMap, posXArray, posXMap, posYArray, posYMap, lines := map[int]int{}, map[int]int{}, []int{}, map[int]int{}, []int{}, map[int]int{}, map[int][]LineItem{}
	for _, pos := range positions {
		tmpXMap[pos[0]]++
		tmpXMap[pos[2]]++
	}
	for k := range tmpXMap {
		posXArray = append(posXArray, k)
	}
	sort.Ints(posXArray)
	for i, pos := range posXArray {
		posXMap[pos] = i
	}

	for _, pos := range positions {
		tmpYMap[pos[1]]++
		tmpYMap[pos[3]]++
		tmp1 := lines[posXMap[pos[0]]]
		tmp1 = append(tmp1, LineItem{start: pos[1], end: pos[3], state: 1})
		lines[posXMap[pos[0]]] = tmp1
		tmp2 := lines[posXMap[pos[2]]]
		tmp2 = append(tmp2, LineItem{start: pos[1], end: pos[3], state: -1})
		lines[posXMap[pos[2]]] = tmp2
	}
	for k := range tmpYMap {
		posYArray = append(posYArray, k)
	}
	sort.Ints(posYArray)
	for i, pos := range posYArray {
		posYMap[pos] = i
	}
	return posXMap, posXArray, posYMap, posYArray, lines
}

// LineItem define
type LineItem struct { // 垂直于 x 轴的线段
	start, end, state int // state = 1 代表进入，-1 代表离开
}

// SegmentItem define
type SegmentItem struct {
	count int
	val   int
}

// SegmentAreaTree define
type SegmentAreaTree struct {
	data        []int
	tree        []SegmentItem
	left, right int
	merge       func(i, j int) int
}

// Init define
func (sat *SegmentAreaTree) Init(nums []int, oper func(i, j int) int) {
	sat.merge = oper
	data, tree := make([]int, len(nums)), make([]SegmentItem, 4*len(nums))
	for i := 0; i < len(nums); i++ {
		data[i] = nums[i]
	}
	sat.data, sat.tree = data, tree
	if len(nums) > 0 {
		sat.buildSegmentTree(0, 0, len(nums)-1)
	}
}

// 在 treeIndex 的位置创建 [left....right] 区间的线段树
func (sat *SegmentAreaTree) buildSegmentTree(treeIndex, left, right int) {
	if left == right-1 {
		sat.tree[treeIndex] = SegmentItem{count: 0, val: sat.data[left]}
		return
	}
	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
	sat.buildSegmentTree(leftTreeIndex, left, midTreeIndex)
	sat.buildSegmentTree(rightTreeIndex, midTreeIndex, right)
	sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
}

func (sat *SegmentAreaTree) pushUp(treeIndex, leftTreeIndex, rightTreeIndex int) {
	newCount, newValue := sat.merge(sat.tree[leftTreeIndex].count, sat.tree[rightTreeIndex].count), 0
	if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count > 0 {
		newValue = sat.merge(sat.tree[leftTreeIndex].val, sat.tree[rightTreeIndex].val)
	} else if sat.tree[leftTreeIndex].count > 0 && sat.tree[rightTreeIndex].count == 0 {
		newValue = sat.tree[leftTreeIndex].val
	} else if sat.tree[leftTreeIndex].count == 0 && sat.tree[rightTreeIndex].count > 0 {
		newValue = sat.tree[rightTreeIndex].val
	}
	sat.tree[treeIndex] = SegmentItem{count: newCount, val: newValue}
}

func (sat *SegmentAreaTree) leftChild(index int) int {
	return 2*index + 1
}

func (sat *SegmentAreaTree) rightChild(index int) int {
	return 2*index + 2
}

// 查询 [left....right] 区间内的值

// Query define
func (sat *SegmentAreaTree) Query(left, right int) int {
	if len(sat.data) > 0 {
		return sat.queryInTree(0, 0, len(sat.data)-1, left, right)
	}
	return 0
}

func (sat *SegmentAreaTree) queryInTree(treeIndex, left, right, queryLeft, queryRight int) int {
	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
	if left > queryRight || right < queryLeft { // segment completely outside range
		return 0 // represents a null node
	}
	if queryLeft <= left && queryRight >= right { // segment completely inside range
		if sat.tree[treeIndex].count > 0 {
			return sat.tree[treeIndex].val
		}
		return 0
	}
	if queryLeft > midTreeIndex {
		return sat.queryInTree(rightTreeIndex, midTreeIndex, right, queryLeft, queryRight)
	} else if queryRight <= midTreeIndex {
		return sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, queryRight)
	}
	// merge query results
	return sat.merge(sat.queryInTree(leftTreeIndex, left, midTreeIndex, queryLeft, midTreeIndex),
		sat.queryInTree(rightTreeIndex, midTreeIndex, right, midTreeIndex, queryRight))
}

// Update define
func (sat *SegmentAreaTree) Update(updateLeft, updateRight, val int) {
	if len(sat.data) > 0 {
		sat.updateInTree(0, 0, len(sat.data)-1, updateLeft, updateRight, val)
	}
}

func (sat *SegmentAreaTree) updateInTree(treeIndex, left, right, updateLeft, updateRight, val int) {
	midTreeIndex, leftTreeIndex, rightTreeIndex := left+(right-left)>>1, sat.leftChild(treeIndex), sat.rightChild(treeIndex)
	if left > right || left >= updateRight || right <= updateLeft { // 由于叶子节点的区间不在是 left == right 所以这里判断需要增加等号的判断
		return // out of range. escape.
	}

	if updateLeft <= left && right <= updateRight { // segment is fully within update range
		if left == right-1 {
			sat.tree[treeIndex].count = sat.merge(sat.tree[treeIndex].count, val)
		}
		if left != right-1 { // update lazy[] for children
			sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
			sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
			sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
		}
		return
	}
	sat.updateInTree(leftTreeIndex, left, midTreeIndex, updateLeft, updateRight, val)
	sat.updateInTree(rightTreeIndex, midTreeIndex, right, updateLeft, updateRight, val)
	// merge updates
	sat.pushUp(treeIndex, leftTreeIndex, rightTreeIndex)
}