POJ 1151 Atlantis 线段树扫描线

题目：http://poj.org/problem?id=1151

题意：给出n个矩形，给出的方式为给出矩形的左下角和右上角两个点，问这些矩形覆盖的面积

思路：线段树扫描线第一题，留个模板

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;

const int N = 210;
struct line
{
double x, y1, y2;
int f;
}arr
;
struct node
{
int l, r, c;
double len, lf, rf;
}s[N*4];
double y
;
int n, cas;

bool cmp(line a, line b)
{
return a.x < b.x;
}
void build(int l, int r, int k)
{
s[k].l = l, s[k].r = r;
s[k].c = s[k].len = 0;
s[k].lf = y[l], s[k].rf = y[r];
if(l + 1 == r) return;
int mid = (l + r) >> 1;
build(l, mid, k << 1);
build(mid, r, k << 1|1);
}
void cal(int k)
{
if(s[k].c > 0) s[k].len = s[k].rf - s[k].lf;
else
{
if(s[k].l + 1 == s[k].r) s[k].len = 0;
else s[k].len = s[k<<1].len + s[k<<1|1].len;
}
}
void update(line a, int k)
{
if(a.y1 == s[k].lf && a.y2 == s[k].rf)
{
s[k].c += a.f;
cal(k);
return;
}
if(a.y2 <= s[k<<1].rf) update(a, k << 1);
else if(a.y1 >= s[k<<1|1].lf) update(a, k << 1|1);
else
{
line tmp = a;
tmp.y2 = s[k<<1].rf;
update(tmp, k << 1);
tmp = a;
tmp.y1 = s[k<<1|1].lf;
update(tmp, k << 1|1);
}
cal(k);
}
int main()
{
double x1, y1, x2, y2;
while(scanf("%d", &n), n)
{
int k = 1;
for(int i = 1; i <= n; i++)
{
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
arr[k].x = x1, arr[k].y1 = y1, arr[k].y2 = y2, arr[k].f = 1, y[k++] = y1;
arr[k].x = x2, arr[k].y1 = y1, arr[k].y2 = y2, arr[k].f = -1, y[k++] = y2;
}
sort(arr + 1, arr + k, cmp);
sort(y + 1, y + k);
build(1, k - 1, 1);
update(arr[1], 1);
double res = 0.0;
for(int i = 2; i < k; i++)
{
res += s[1].len * (arr[i].x - arr[i-1].x);
update(arr[i], 1);
}
printf("Test case #%d\nTotal explored area: %.2f\n\n", ++cas, res);
}
return 0;
}