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给定若干矩形，求被奇数个矩形覆盖到的面积总和。这个问题与传统的矩形覆盖问题不同，传统的问题通常是计算被完全覆盖的区域面积总和，而这里是奇数次覆盖的区域。为了解决这个问题，可以采用线段树和扫描线算法结合懒标记技术。

思路：扫描线

为了处理这个问题，采用扫描线算法，结合线段树和懒标记技术。扫描线算法通过离散化处理，将矩形的左右端点映射到线段树的区间，这样可以高效地处理区间更新和查询操作。

用到的小算法：离散化 + 线段树

问题分析

• 传统的扫描线算法在处理完所有矩形后，总区间的覆盖次数会归零，但在这个问题中，可能存在剩余区间未被覆盖。因此，需要使用懒标记来维护这些区间的状态。
• 需要处理多个矩形部分重叠的情况，确保懒标记被正确传递，维护子树的状态。

解决方案

代码实现

import sys
def main():
    sys.setrecursionlimit(1 << 25)
    n = int(sys.stdin.readline())
    points = []
    lines = []
    for _ in range(n):
        x1, y1, x2, y2 = map(int, sys.stdin.readline().split())
        points.append(x1)
        points.append(x2)
        lines.append((x1, x2, y1, 1))
        lines.append((x1, x2, y2, -1))
    
    points = sorted(list(set(points)))
    lines = sorted(lines, key=lambda x: (x[0], x[1]))
    len_points = len(points)
    xx = [0] * (2 * n + 1)
    for i in range(2 * n):
        xx[i] = points[i]
    
    tree = [0] * (4 * len(points))
    for i in range(len(points)):
        tree[i] = {'l': points[i], 'r': points[i], 'len': 0, 'cover': 0}
    
    def pushup(p):
        if p >= len(tree) or p < 0:
            return
        tree[p]['len'] = tree[left(p)]['len'] + tree[right(p)]['len']
    
    def down(p, l, r):
        if tree[p]['cover'] == 0:
            return
        mid = (l + r) // 2
        if l == r:
            return
        left_p = left(p)
        right_p = right(p)
        tree[left_p]['cover'] ^= 1
        tree[right_p]['cover'] ^= 1
        len_left = mid - left(p) + 1
        len_right = r - mid
        tree[left_p]['len'] = len_left - tree[left_p]['len']
        tree[right_p]['len'] = len_right - tree[right_p]['len']
        tree[p]['cover'] = 0
    
    def build(p, l, r):
        if l == r:
            return
        mid = (l + r) // 2
        build(left(p), l, mid)
        build(right(p), mid + 1, r)
        pushup(p)
    
    def update(p, l, r, f):
        if tree[p]['l'] >= l and tree[p]['r'] <= r:
            tree[p]['cover'] ^= 1
            tree[p]['len'] = (tree[p]['r'] - tree[p]['l'] + 1) - tree[p]['len']
            return
        down(p, tree[p]['l'], tree[p]['r'])
        mid = (tree[p]['l'] + tree[p]['r']) // 2
        if l <= mid:
            update(left(p), l, r, f)
        if r > mid:
            update(right(p), l, r, f)
        pushup(p)
    
    build(1, 0, len(points) - 1)
    
    total = 0
    for i in range(2 * n):
        x1, x2, y1, f = lines[i]
        l = 0
        r = len(points) - 1
        while l < r:
            mid = (l + r) // 2
            if lines[i][0] < points[mid]:
                l = mid + 1
            else:
                r = mid
        left_idx = l
        right_idx = r
        if right_idx >= len(points):
            right_idx = len(points) - 1
        while left_idx < right_idx:
            mid = (left_idx + right_idx) // 2
            if points[mid] < x1:
                left_idx = mid + 1
            else:
                right_idx = mid
        left_idx = left_idx if points[left_idx] <= x1 else left_idx + 1
        right_idx = right_idx if points[right_idx] >= x2 else right_idx - 1
        if left_idx >= right_idx:
            continue
        if lines[i][3] == 1:
            y = y1
        else:
            y = y2
        if i == 0:
            h = y - 0
        else:
            h = y - lines[i-1][2]
        total += tree[1]['len'] * h
        update(1, left_idx, right_idx, f)
    
    print(total)
    
if __name__ == "__main__":
    main()

代码解释

通过以上步骤，可以高效地计算被奇数次覆盖的区域面积总和。

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