hdu 1255(离散化+扫描线)

题意：有n个矩形，给出每个矩形左下角坐标和右上角坐标，求矩形覆盖两次以上的总面积。

题解：因为要覆盖两次以上，线段树要维护两个值，覆盖一次的长度和覆盖两次的长度，在区间合并的时候，如果标记是覆盖大于一次，那么覆盖一次的长度为0，覆盖两次的长度为区间长度，标记是覆盖一次，那么覆盖两次的长度是左右子区间覆盖一次和覆盖两次的和，而覆盖一次的就是区间长度减覆盖两次，如果标记为0，就是覆盖一次和覆盖两次是各自左右子区间的和，叶子节点直接赋0。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
const int N = 2005;
struct Line {
double lx, rx, h;
int flag;
Line(double a, double b, double c, int d):lx(a), rx(b), h(c), flag(d) {}
bool operator < (const Line &a) const { return h < a.h; }
};
int n, flag[N << 2];
double tree[3][N << 2];
vector<double> a;
vector<Line> line;
map<double, int> mp;

void pushup(int k, int left, int right) {
if (flag[k] > 1) {
tree[2][k] = tree[0][k];
tree[1][k] = 0;
}
else if (flag[k] == 1) {
if (left + 1 == right)
tree[2][k] = 0;
else
tree[2][k] = tree[2][k * 2] + tree[2][k * 2 + 1] + tree[1][k * 2] + tree[1][k * 2 + 1];
tree[1][k] = tree[0][k] - tree[2][k];
}
else {
if (left + 1 == right)
tree[1][k] = tree[2][k] = 0;
else {
tree[1][k] = tree[1][k * 2] + tree[1][k * 2 + 1];
tree[2][k] = tree[2][k * 2] + tree[2][k * 2 + 1];
}
}
}

void build(int k, int left, int right) {
tree[0][k] = a[right] - a[left];
tree[1][k] = tree[2][k] = flag[k] = 0;
if (left + 1 != right) {
int mid = (left + right) / 2;
build(k * 2, left, mid);
build(k * 2 + 1, mid, right);
}
}

void modify(int k, int left, int right, int l, int r, int v) {
if (l <= left && right <= r)  {
flag[k] += v;
pushup(k, left, right);
return;
}
int mid = (left + right) / 2;
if (l < mid)
modify(k * 2, left, mid, l, r, v);
if (r > mid)
modify(k * 2 + 1, mid, right, l, r, v);
pushup(k, left, right);
}

int main() {
int t;
scanf("%d", &t);
while (t--) {
a.clear(), line.clear(), mp.clear();
scanf("%d", &n);
double x1, y1, x2, y2;
for (int i = 1; i <= n; i++) {
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
line.push_back(Line(x1, x2, y1, 1));
line.push_back(Line(x1, x2, y2, -1));
a.push_back(x1);
a.push_back(x2);
}
sort(line.begin(), line.end());
sort(a.begin(), a.end());
a.erase(unique(a.begin(), a.end()), a.end());
int sz = a.size(), sz2 = line.size();
for (int i = 0; i < sz; i++)
mp[a[i]] = i;
build(1, 0, sz - 1);
double res = 0;
for (int i = 0; i < sz2; i++) {
if (i != 0)
res += (line[i].h - line[i - 1].h) * tree[2][1];
modify(1, 0, sz - 1, mp[line[i].lx], mp[line[i].rx], line[i].flag);
}
printf("%.2lf\n", res);
}
return 0;
}