计蒜客 Overlapping Rectangles （离散化）

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 2000 + 7;
struct R{
    int bx, by, tx, ty;
}rec[maxn];

int n;
int nx = 0 , ny = 0;
int X[maxn], Y[maxn];
bool cover[maxn][maxn];
int solve(){
    sort(X,X+nx); sort(Y,Y+ny);
    int ux = unique(X,X+nx) - X;//排序，去重，用点表示线段， 就是离散化的核心
    int uy = unique(Y,Y+ny) - Y;

//    cout << ux << " " << uy << "\n";

    for(int i = 0; i < n ; i++){
        int tx = lower_bound(X, X+ux, rec[i].bx) - X; // 找到左下角x对应的离散化点
        int ty = lower_bound(Y, Y+uy, rec[i].by) - Y; // 找到左下角y对应的离散化点

        for(int j = tx ; X[j] < rec[i].tx; j++){//calculate the area have covered.
            for(int k = ty; Y[k] < rec[i].ty; k++){
                cover[j][k] = 1; //cover数组表示哪些离散化点是属于矩形内的
            }
        }
    }

    int area = 0;
    for(int i = 0; i < ux; i++){
        for(int j = 0; j < uy; j++){
            if(cover[i][j]){
                area += (X[i+1]-X[i]) * (Y[j+1]-Y[j]);
            }
        }
    }
    return area;

}
int main(){
//    freopen("data.txt","r", stdin);
    while(~scanf("%d", &n)){
        memset(cover,0,sizeof(cover));
        nx = 0 , ny = 0;
        if(n == 0){
            printf("*\n");
            break;
        }
        for(int i = 0; i < n; i++){
            scanf("%d %d %d %d", &rec[i].bx, &rec[i].by, &rec[i].tx, &rec[i].ty);
            X[nx++] = rec[i].bx, X[nx++] = rec[i].tx;
            Y[ny++] = rec[i].by, Y[ny++] = rec[i].ty;
        }
        printf("%d\n",solve());

    }
}