HDU 1542 Atlantis(线段树矩形覆盖)

题意：给出若干个矩形，求出总共的面积

思路：线段树 + 扫描线，从下而上扫描，遇到矩形下边+1（插入新的覆盖线段），否则-1（删除该线段），求出每次覆盖总长度乘以每次的高度差，加起来就是答案

#include<cstdio>
#include<cstring>
#include<algorithm>
const int maxn = 2 * 1e2 + 10;
using namespace std;

struct mt {
double l, r, h;
int add;
mt() {}
mt(double l, double r, double h, int a) :
l(l), r(r), h(h), add(a) {}
bool operator < (mt p) const { return h < p.h; }
} Mt[2 * maxn];
double x[2 * maxn], seg[4 * maxn];
int cover[4 * maxn];
int n, ql, qr, kase = 1;
double X1, X2, Y1, Y2;

void update(int node, int l, int r, int add) {
if(l >= ql && r <= qr) {
cover[node] += add;
if(cover[node]) seg[node] = x[r + 1] - x[l];
else if(l == r) seg[node] = 0;
else seg[node] = seg[2 * node] + seg[2 * node + 1];
return ;
}
int mid = (l + r) >> 1;
if(ql <= mid) update(node * 2, l, mid, add);
if(qr > mid) update(node * 2 + 1, mid + 1, r, add);
if(cover[node]) seg[node] = x[r + 1] - x[l];
else seg[node] = seg[node * 2] + seg[node * 2 + 1];
}

int main() {
while(scanf("%d", &n) && n) {
memset(cover, 0, sizeof(cover));
memset(seg, 0, sizeof(seg));
int num = 0;
for(int i = 0; i < n; i++) {
scanf("%lf %lf %lf %lf", &X1, &Y1, &X2, &Y2);
Mt[num] = mt(X1, X2, Y1, 1);
Mt[num + 1] = mt(X1, X2, Y2, -1);
x[num] = X1; x[num + 1] = X2;
num += 2;
}
sort(x, x + num);
sort(Mt, Mt + num);
int nn = unique(x, x + num) - x;
double ans = 0;
for(int i = 0; i < num - 1; i++) {
ql = lower_bound(x, x + nn, Mt[i].l) - x;
qr = lower_bound(x, x + nn, Mt[i].r) - x - 1;
update(1, 0, num - 1, Mt[i].add);
ans += (Mt[i + 1].h - Mt[i].h) * seg[1];
}
printf("Test case #%d\n", kase++);
printf("Total explored area: %.2f\n\n", ans);
}
return 0;
}