P1884 [USACO12FEB]Overplanting S——洛谷

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;
}