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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。
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。
import sysdef 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() 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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