探讨了判断多个矩形是否能精确覆盖一个大矩形的问题，提供了三种解法：暴力检查重叠与面积总和，四叉树算法，及利用矩形顶点计数的巧妙方法。
我们有 N 个与坐标轴对齐的矩形, 其中 N > 0, 判断它们是否能精确地覆盖一个矩形区域。
每个矩形用左下角的点和右上角的点的坐标来表示。例如， 一个单位正方形可以表示为 [1,1,2,2]。 ( 左下角的点的坐标为 (1, 1) 以及右上角的点的坐标为 (2, 2) )。
示例 1:
rectangles = [ [1,1,3,3], [3,1,4,2], [3,2,4,4], [1,3,2,4], [2,3,3,4] ] 返回 true。5个矩形一起可以精确地覆盖一个矩形区域。
示例 2:
rectangles = [ [1,1,2,3], [1,3,2,4], [3,1,4,2], [3,2,4,4] ] 返回 false。两个矩形之间有间隔，无法覆盖成一个矩形。
示例 3:
rectangles = [ [1,1,3,3], [3,1,4,2], [1,3,2,4], [3,2,4,4] ] 返回 false。图形顶端留有间隔，无法覆盖成一个矩形。
示例 4:
rectangles = [ [1,1,3,3], [3,1,4,2], [1,3,2,4], [2,2,4,4] ] 返回 false。因为中间有相交区域，虽然形成了矩形，但不是精确覆盖。
这道题我在写前两个解法的时候还没ac通过，但是已经忍不住要写了。思路还是很清晰的，遍历矩形，算有没有重合，面积累加，最后看看总面积是不是最大范围的覆盖的矩形的面积。
第一遍暴力解法：
class Solution { public: bool Cover(vector<int>& a, vector<int>&b) { if (min(a[0], a[2]) >= max(b[0], b[2]) || max(a[0], a[2]) <= min(b[0], b[2]) || min(a[1], a[3]) >= max(b[1], b[3]) || max(a[1], a[3]) <= min(b[1], b[3])) { return false; } return true; } bool isRectangleCover(vector<vector<int>>& rectangles) { int minx, miny, maxx, maxy; minx = miny = INT_MAX; maxx = maxy = INT_MIN; long long int sum = 0; for (int i = 0; i < rectangles.size();i++) { for (int r = 0; r < i;r++) { if (Cover(rectangles[i], rectangles[r])) { return false; } } minx = min(minx, min(rectangles[i][0], rectangles[i][2])); miny = min(miny, min(rectangles[i][1], rectangles[i][3])); maxx = max(maxx, max(rectangles[i][0], rectangles[i][2])); maxy = max(maxy, max(rectangles[i][1], rectangles[i][3])); sum += abs((rectangles[i][0] - rectangles[i][2])*(rectangles[i][1] - rectangles[i][3])); } return sum == (maxx - minx)*(maxy - miny); } };
果不其然挂了，测试最后两组都是3w+的数据量
第二遍：四叉树解法
class QuadNode { public: enum{ quad_1,//四个象限 quad_2, quad_3, quad_4, quad_count, }; vector<vector<int>>* data; QuadNode* Children[quad_count];//孩子指针,数组大小为8 QuadNode* Parent;//父节点指针 typedef std::list<int> RecList; typedef std::list<int>::iterator RecListIter; RecList rectlist;//携带的参数 实体列表 int quad;//在父节点中的象限 int deep;//自己所在的层索引 int minx,miny; int maxx,maxy; QuadNode(vector<vector<int>>* data, int x1, int x2, int y1, int y2, int dp, int qd) { minx = x1; maxx = x2; miny = y1; maxy = y2; deep = dp; quad = qd; Parent = NULL; this->data = data; memset(Children, 0, sizeof(Children)); } ~QuadNode() { for (int i = 0; i < quad_count; i++) { if (Children[i]) { delete Children[i]; Children[i] = NULL; } } rectlist.clear(); } QuadNode* GetDeepest(int index) { if (deep > 0) { //4个孩子都要创建 for (int r = 0; r < QuadNode::quad_count; r++) { if (!Children[r]) { int ix = r == QuadNode::quad_1 || r == QuadNode::quad_3 ? minx : (minx + maxx) / 2; int ax = r == QuadNode::quad_1 || r == QuadNode::quad_3 ? (minx + maxx) / 2 : maxx; int iy = r == QuadNode::quad_1 || r == QuadNode::quad_2 ? miny : (miny + maxy) / 2; int ay = r == QuadNode::quad_1 || r == QuadNode::quad_2 ? (miny + maxy) / 2 : maxy; QuadNode *node = new QuadNode(data, ix, ax, iy, ay, deep - 1, r); node->Parent = this; Children[r] = node; } if (Children[r]->CheckInRange(index)) { return Children[r]->GetDeepest(index); } } } return this; } bool CheckInRange(int index) { if ((*data)[index][0] >= minx && (*data)[index][2] <= maxx && (*data)[index][1] >= miny && (*data)[index][3] <= maxy) { return true; } return false; } bool CheckCover(int index) { QuadNode* n = GetDeepest(index); QuadNode* parent = n->Parent; while (parent) { if (parent->CheckWithTrianglelist(index)) { return true; } parent = parent->Parent; } if (n->CollisionCheck(index)) return true; n->rectlist.push_back(index); return false; } bool CollisionCheck(int index) { return CheckWithTrianglelist(index) || CheckWithSubSpace(index); } bool CheckWithTrianglelist(int index) { RecListIter itr = rectlist.begin(); while (itr != rectlist.end()) { int id = *itr; if (Cover((*data)[id], (*data)[index])) { return true; } itr++; } return false; } bool CheckWithSubSpace(int index) { bool collision = false; for (int i = 0; i < quad_count && Children[i]; i++) { int vec[] = { minx, miny, maxx, maxy }; vector<int> para(vec, vec + 4); if (Cover((*data)[index], para)) { collision |= Children[i]->CollisionCheck(index); } if (collision) { return true; } } return false; } bool Cover(vector<int>& a, vector<int>&b) { if (min(a[0], a[2]) >= max(b[0], b[2]) || max(a[0], a[2]) <= min(b[0], b[2]) || min(a[1], a[3]) >= max(b[1], b[3]) || max(a[1], a[3]) <= min(b[1], b[3])) { return false; } return true; } }; class Solution { public: int GetMax2Power(int xmax, int ymax, int& lg) { int max = xmax; if (ymax > max) max = ymax; if ((max & (max - 1)) == 0) { double L = log(max*1.0) / log(2.0); lg = (int)L + 1; return max; } else { double L = log(max*1.0) / log(2.0); lg = (int)L + 2; return (int)pow(2 * 1.0, lg - 1.0); } } bool isRectangleCover(vector<vector<int>>& rectangles) { int minx, miny, maxx, maxy; minx = miny = INT_MAX; maxx = maxy = INT_MIN; long long int sum = 0; for (int i = 0; i < rectangles.size(); i++) { minx = min(minx, min(rectangles[i][0], rectangles[i][2])); miny = min(miny, min(rectangles[i][1], rectangles[i][3])); maxx = max(maxx, max(rectangles[i][0], rectangles[i][2])); maxy = max(maxy, max(rectangles[i][1], rectangles[i][3])); } int mx = max(abs(maxx), abs(minx)); int my = max(abs(maxy), abs(miny)); int range, lg; range = GetMax2Power(mx, my, lg); //四叉树 QuadNode* root = new QuadNode(&rectangles, -range, range, -range, range, lg, 0); for (int i = 0; i < rectangles.size();i++) { if (root->CheckCover(i)) { return false; } sum += abs((rectangles[i][0] - rectangles[i][2])*(rectangles[i][1] - rectangles[i][3])); } delete root; return sum == (maxx - minx)*(maxy - miny); } };
以为没什么问题了，跑一下又超时了，真是恶心的一p啊。拿测试数据跑一跑发现，几乎95%的数据都在边界上，四叉树无法往下细化。写个四叉树容易吗？？？
只能硬着头皮继续想了,果然没有想到。参考了下别人的歪门邪道，感觉前面写的东西都白瞎了。有时候解决问题，还得靠技巧。
解法三：所有的矩形顶点，有且只有四个边角是只出现一次，剩下的顶点要么两次，要么四次
long long int getHash(int x, int y) { long long int t = 2 << 16; return x*t + y; } bool isRectangleCover(vector<vector<int>>& rectangles) { int minx, miny, maxx, maxy; minx = miny = INT_MAX; maxx = maxy = INT_MIN; long long int sum = 0; unordered_set<long long int> st; for (int i = 0; i < rectangles.size(); i++) { minx = min(minx, min(rectangles[i][0], rectangles[i][2])); miny = min(miny, min(rectangles[i][1], rectangles[i][3])); maxx = max(maxx, max(rectangles[i][0], rectangles[i][2])); maxy = max(maxy, max(rectangles[i][1], rectangles[i][3])); sum += abs((rectangles[i][0] - rectangles[i][2])*(rectangles[i][1] - rectangles[i][3])); long long int lu = getHash(rectangles[i][0], rectangles[i][3]); long long int ld = getHash(rectangles[i][0], rectangles[i][1]); long long int ru = getHash(rectangles[i][2], rectangles[i][3]); long long int rd = getHash(rectangles[i][2], rectangles[i][1]); if (st.count(lu) == 0) st.insert(lu); else st.erase(lu); if (st.count(ld) == 0) st.insert(ld); else st.erase(ld); if (st.count(ru) == 0) st.insert(ru); else st.erase(ru); if (st.count(rd) == 0) st.insert(rd); else st.erase(rd); } return sum == (maxx - minx)*(maxy - miny) && st.size() == 4 && st.count(getHash(minx, miny)) == 1 && st.count(getHash(minx, maxy)) == 1 && st.count(getHash(maxx, miny)) == 1 && st.count(getHash(maxx, maxy)) == 1; }
吐血！
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