HDU-1542 Atlanta calculates rectangular area and

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1542

This question is given n rectangles, and the area after the superposition is obtained. It is similar to the area intersection, that is, there is no requirement for the number of covers.

The Code is as follows:

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
#include <algorithm>
#define MAXN 205
using namespace std;

struct Node
{
double x, y1, y2;
int lr;
bool operator < (Node t) const
    {
return t.x-x>1e-6;
    }
}e[MAXN*2];

struct
{
int l, r;
int cover;
}seg[MAXN*3];

double yy[MAXN*2];

void creat(int f, int l, int r)
{
int mid = (l+r)>>1;
    seg[f].l = l, seg[f].r = r;
    seg[f].cover = 0;
if (r - l > 1)
    {
        creat(f<<1, l, mid);
        creat(f<<1|1, mid, r);
    }
}

void modify(int f, int l, int r, int val)
{
int mid = (seg[f].l + seg[f].r) >> 1;
if (seg[f].l == l && seg[f].r == r)
    {
        seg[f].cover += val;
    }
else if (seg[f].r - seg[f].l > 1)
    {
if (r <= mid)
            modify(f<<1, l, r, val);
else if (l >= mid)
            modify(f<<1|1, l, r, val);
else
        {
            modify(f<<1, l, mid, val);
            modify(f<<1|1, mid, r, val);
        }
    }
}

void query(int f, double &ans)
{
if (seg[f].cover > 0)
    {
        ans += yy[seg[f].r] - yy[seg[f].l];
    }
else if (seg[f].r - seg[f].l > 1)
    {
        query(f<<1, ans);
        query(f<<1|1, ans);
    }
}

int main()
{
double x1, y1, x2, y2, ans, res;
int N, ca = 1;;
while (scanf("%d", &N), N)
    {
        res = 0;
        map<double,int>mp;
for (int i = 1, j = 1; i <= N; ++i, j += 2)
        {
            scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
            e[j].x = x1, e[j].y1 = y1, e[j].y2 = y2; 
            e[j].lr = 1;
            e[j+1].x = x2, e[j+1].y1 = y1, e[j+1].y2 = y2;
            e[j+1].lr = -1;
            yy[j] = y1, yy[j+1] = y2;
        }
        sort(e+1, e+1+2*N);
        sort(yy+1, yy+1+2*N);
int cnt = unique(yy+1, yy+1+2*N) - (yy+1);
        creat(1, 1, cnt);
for (int i = 1; i <= cnt; ++i)
        {
            mp[yy[i]] = i;
//    printf("yy= %lf\n", yy[i]);
        }
for (int i = 1; i < 2*N; ++i)
        {
            ans = 0;
            modify(1, mp[e[i].y1], mp[e[i].y2], e[i].lr);
            query(1, ans);
            res += ans * (e[i+1].x - e[i].x);
//    printf("res = %lf\n", res);
        }
        printf("Test case #%d\nTotal explored area: %.2lf\n\n", ca++, res);
    } 
return 0;
}