Problem Dscription
Give you n rectangles, each rectangle gives you the lower left subscript and the lower right upper corner, let you find the area intersection.
Train of thought:
Reference Blog: http://www.cnblogs.com/scau20110726/archive/2013/04/14/3020998.html It is suggested that we do hdu 1542 first and then look at this question. The code was only slightly changed.
The segment tree maintains len2 of cover >= 2, len of cover > 0, and whether the whole segment of the interval is valid or not. Compared with hdu 1542, there is an extra len2. So get_len2() function has changed, specific details refer to the above blog, very detailed.
#include<bits/stdc++.h>
using namespace std;
#define lson root<<1
#define rson root<<1|1
#define N 1005
#define MID int mid = (l + r) / 2
struct node
{
int cover;
double len, len2;//There are more len2 in different places.
};
node tree[N<<3];
double yy[N<<2];
struct Node
{
int cover;
double x;
double yd, yu;
bool operator < (const Node &b) const{
return x < b.x;
}
};
Node a[N<<2];
void build(int root, int l, int r)
{
tree[root].cover = tree[root].len = tree[root].len2 = 0;
if(l + 1 == r)
return;
MID;
build(lson, l, mid);
build(rson, mid, r);
}
void get_len2(int root, int l, int r)
{
if(tree[root].cover)
tree[root].len = yy[r] - yy[l];
else if(l + 1 == r)
tree[root].len = 0;
else tree[root].len = tree[lson].len + tree[rson].len;
//Here are some differences
if(tree[root].cover > 1) tree[root].len2 = yy[r] - yy[l];//Cove >= 2 interval length is valid
else if(l + 1 == r) tree[root].len2 = 0;//Leaf nodes, and the whole section of the interval is not valid
else if(tree[root].cover == 1)//With len of left and right children, len 2 will be satisfied.
tree[root].len2 = tree[lson].len + tree[rson].len;
else tree[root].len2 = tree[lson].len2 + tree[rson].len2;//The sum of left and right children len2
//
}
void updata(int root, int l, int r, int ul, int ur, int cover)
{
if(ul <= l && r <= ur)
{
tree[root].cover += cover;
get_len2(root, l, r);
return;
}
MID;
if(ul < mid) updata(lson, l, mid, ul, ur, cover);
if(ur > mid) updata(rson, mid, r, ul, ur, cover);
get_len2(root, l, r);
}
int main()
{
int n, cnt, i, T;
double x1, x2, y1, y2;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
cnt = 0;
for(i = 0; i < n; i++)
{
scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
yy[cnt] = y1;
a[cnt++] = ((Node){1, x1, y1, y2});
yy[cnt] = y2;
a[cnt++] = ((Node){-1, x2, y1, y2});
}
sort(yy, yy + cnt);
n = unique(yy, yy + cnt) - yy;
build(1, 0, n - 1);
sort(a, a + cnt);
int ul, ur;
double ans = 0;
for(i = 0; i < cnt - 1; i++)
{
ul = lower_bound(yy, yy + n, a[i].yd) - yy;
ur = lower_bound(yy, yy + n, a[i].yu) - yy;
updata(1, 0, n - 1, ul, ur, a[i].cover);
ans += tree[1].len2 * (a[i + 1].x - a[i].x);
}
printf("%.2lf\n", ans);
}
}