经典算法 | 给定n个矩形，判断这些矩形是否在不重合的情况下组成一个大矩形的算法

GivenN axis-aligned rectangles where N > 0, determine if they all together forman exact cover of a rectangular region.
Eachrectangle is represented as a bottom-left point and a top-right point. Forexample, a unit square is represented as [1,1,2,2]. (coordinate of bottom-leftpoint is (1, 1) and top-right point is (2, 2)).
Example1:

rectangles = [

  [1,1,3,3],

  [3,1,4,2],

  [3,2,4,4],

  [1,3,2,4],

  [2,3,3,4]

]

 

Return true. All 5 rectangles together form an exact cover ofa rectangular region.

Example2:

rectangles = [

  [1,1,2,3],

  [1,3,2,4],

  [3,1,4,2],

  [3,2,4,4]

]

 

Return false. Because there is a gap between the tworectangular regions.

Example3:

rectangles = [

  [1,1,3,3],

  [3,1,4,2],

  [1,3,2,4],

  [3,2,4,4]

]

 

Return false. Because there is a gap in the top center.

Example4:

rectangles = [

  [1,1,3,3],

  [3,1,4,2],

  [1,3,2,4],

  [2,2,4,4]

]

 

Return false. Because two of the rectangles overlap with eachother.

这一题我也没想到会这么难，这一题要求你对于给定的所有矩形，判断其是否能在不相交的情况下组成一个大矩形。



每个矩形的四个端点在大矩形中只有可能是以上三种情况的一种，



每个矩形有4个端点，并且每个端点在大矩形中有4个区域，当某个端点在这4个区域的某个区域覆盖了超过一个的时候，这个时候必有两个矩形重叠，

所有在大矩形四个端点的端点有且只有一个区域被覆盖，在大矩形四条边上的端点只能有两个区域被覆盖或者是无区域覆盖

在大矩形里面的端点只能有两个相邻区域覆盖或者是四个区域覆盖

对于输入的所有矩形，对其四个端点计算覆盖值，当所有矩形输入完毕之后，统计所有端点的覆盖值是否合法就能得到这种情况的合法性

这个技巧还能用来判断多个矩形组成的大矩形在小矩形两两覆盖合法的情况下是否有空腔的问题，但是不能用来判断多个矩形是否两两重叠

class Solution {
public:
bool setIn(int x, int y, int whatDrt, unordered_map<string, int> &mark)
{
unordered_map<string, int> ::iterator it;
string tmp;
if (x == 3 && y == 1)
int j_mark = 1;

tmp = to_string(x) + "," + to_string(y);
it = mark.find(tmp);
if (it == mark.end())
{
mark.insert(pair <string,int>(tmp, whatDrt));
}
else
{
int t = (*it).second;
if (((*it).second&whatDrt) != 0)
return false;
(*it).second ^= whatDrt;
}
return true;
}
bool putOn(unordered_map<string, int> &mark,int x1,int y1,int x2,int y2)
{
if (!setIn(x1, y1, 2, mark)) return false;
if (!setIn(x1, y2, 8, mark)) return false;
if (!setIn(x2, y1, 1, mark)) return false;
if (!setIn(x2, y2, 4, mark)) return false;
return true;
}
bool isRectangleCover(vector<vector<int>>& rectangles) {
unordered_map <string, int> mark;
string tmp; int x1=INT_MAX, x2=0, y1= INT_MAX, y2=0,count=0;
for (int i = 0; i < rectangles.size(); i++)
{
x1 = min(x1, rectangles[i][0]);
y1 = min(y1, rectangles[i][1]);
x2 = max(x2, rectangles[i][2]);
y2 = max(y2, rectangles[i][3]);
if (!putOn(mark, rectangles[i][0], rectangles[i][1], rectangles[i][2], rectangles[i][3])) return false;
}
string tmp1 = to_string(x1) + ',' + to_string(y1);
string tmp2 = to_string(x1) + ',' + to_string(y2);
string tmp3 = to_string(x2) + ',' + to_string(y1);
string tmp4 = to_string(x2) + ',' + to_string(y2);
unordered_map<string, int> ::iterator it;
for (it = mark.begin(); it != mark.end(); it++)
{
if ((*it).first == tmp1 || (*it).first == tmp2 || (*it).first == tmp3 || (*it).first == tmp4) { count++; continue; }
string x = (*it).first;
if ((*it).second == 14 || (*it).second == 13 || (*it).second == 11 || (*it).second == 7 || (*it).second == 9 || (*it).second == 6) return false;
if ((*it).second == 1 || (*it).second == 2 || (*it).second == 4 || (*it).second == 8) return false;
}
if (count != 4) return false;
return true;
}
};