No.88 - LeetCode850 - 多个矩形重叠后面积

方法一：
扫描线+数组区间更新：时间复杂度O(N*N)

class Solution {
public: 
    struct StateNode{
        int y;
        int tp;
        int x1;
        int x2;
        void puts(int yy,int t,int xa,int xb){
            y = yy; tp = t; x1 = xa; x2 = xb;
        }
        bool operator < (const StateNode& x) const{
            if(this->y < x.y) return true;
            return false;
        }
    };
    int rectangleArea(vector<vector<int>>& rectangles) {
        int N = rectangles.size();
        StateNode sn[N*2];
        map<int,int> mp,pm;
        for(int i=0;i<N;i++){
            sn[i].puts(rectangles[i][1],1,rectangles[i][0],rectangles[i][2]);
            sn[i+N].puts(rectangles[i][3],-1,rectangles[i][0],rectangles[i][2]);
            mp[rectangles[i][0]] = mp[rectangles[i][2]] = 1;
        }
        sort(sn,sn+N+N);
        int M = 1;
        for(map<int,int>::