解题思路： 离散化 + 差分矩阵
解题步骤：

因为原始坐标系与差分的坐标系不同，所以要进行坐标系的变换

离散化处理缩小查找空间

图中坐标离散化后，再转化为格子的坐标，因为之前存贮的坐标是点的坐标，每个格子存储一个矩形

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
const int N = 2020;
vector<int> allx, ally;//存放所有点的横纵坐标
map<int, int> mapx, mapy;//映射坐标对应的下标
vector<PII> rec[2];//存放矩形
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2, int c)
{b[x1][y1] += c;b[x2 + 1][y1] -= c;b[x1][y2 + 1] -= c;b[x2 + 1][y2 + 1] += c;
}
int main()
{int n; cin >> n;//输入所有的x, y 坐标到数组allx和ally中， 准备离散化allx.push_back(0);ally.push_back(0);for(int i = 0; i < n; i++){int x1, y1, x2, y2;cin >> x1 >> y1 >> x2 >> y2;rec[0].push_back({y2, x1}); //注意坐标系的转换rec[1].push_back({y1, x2});allx.push_back(y1);allx.push_back(y2);ally.push_back(x1);ally.push_back(x2);}//排序， 去重sort(allx.begin() + 1, allx.end());allx.erase(unique(allx.begin() + 1, allx.end()), allx.end());sort(ally.begin() + 1, ally.end());ally.erase(unique(ally.begin() + 1, ally.end()), ally.end());//建立x 和y坐标的map映射for(int i = 1; i < allx.size(); i++) mapx[allx[i]] = i;for(int i = 1; i < ally.size(); i++) mapy[ally[i]] = i;//差分矩阵for(int i = 0; i < n; i++){int x1, y1 , x2, y2;x1 = rec[0][i].first;//点的实际坐标y1 = rec[0][i].second;x2 = rec[1][i].first;y2 = rec[1][i].second;insert(mapx[x1], mapy[y1], mapx[x2] - 1, mapy[y2] - 1, 1);//映射为格子下标}//求前缀和矩阵for(int i = 1; i <= allx.size(); i++)for(int j = 1; j <= ally.size(); j++)a[i][j] = a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1] + b[i][j];//将前缀和矩阵中被覆盖的格子，还原成原坐标系中的矩阵，进行面积求和LL area = 0;for(int i = 1; i <= allx.size(); i++)for(int j = 1; j <= ally.size(); j++){if(a[i][j]){int x1, y1, x2, y2;x1 = allx[i];y1 = ally[j];x2 = allx[i + 1];y2 = ally[j + 1];area += (LL)(x2 - x1) * (y2 - y1);}}cout << area << endl;return 0;
}