HDU - 1255 覆盖的面积(线段树 扫描线)

题目大意：中文题

解题思路：记录区间被cover的次数，区间的被cover1次的长度len，区间被cover2次或者2次以上的长度twice

所要求的覆盖面积，就是求被cover两次或者两次以上的长度 * 高的和

首先判断是否被cover过，先更新一下len

1.接着判断一下是否被cover过次或者两次以上，如果符合，直接等于右端点-左端点

2.如果不符合条件1，且l == r ，那么twice = 0

3.如果不符合2，但是满足cover == 1，那么twice[u] = len[u << 1] + len[u << 1 | 1]， 因为左右区间被覆盖一次的长度会被再覆盖一次

4.如果不符合条件3，twice[u] = twice[u << 1] + twice[u << 1 | 1]

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 2010;
const int M = 2010 << 2;

struct Segment{
double l, r, h;
int f;
}S
;

int cnt, n, m;
int cover[M];
double pos
, len[M], twice[M];

void build(int u, int l, int r) {
len[u] = twice[u] = 0;
cover[u] = 0;
if (l == r) return ;
int mid = (l + r) >> 1;
build(u << 1, l, mid);
build(u << 1 | 1, mid + 1, r);
}

int cmp(const Segment a, const Segment b) {
return a.h < b.h;
}

void init() {
scanf("%d", &n);
double x1, y1, x2, y2;

cnt = 1;
for (int i = 1; i <= n; i++) {
scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
pos[cnt] = S[cnt].l = S[cnt + 1].l = x1;
pos[cnt + 1] = S[cnt].r = S[cnt + 1].r = x2;
S[cnt].h = y1; S[cnt + 1].h = y2;
S[cnt].f = 1; S[cnt + 1].f = -1;
cnt += 2;
}
sort(S + 1, S + cnt, cmp);
sort(pos + 1, pos + cnt);

m = 1;
for (int i = 2; i < cnt; i++)
if (pos[i] != pos[i - 1]) pos[++m] = pos[i];
build(1, 1, m);
}

void getlen(int u, int l, int r) {
if (cover[u]) len[u] = pos[r + 1] - pos[l];
else if (l ==  r) len[u] = 0;
else len[u] = len[u << 1] + len[u << 1 | 1];

if (cover[u] > 1) twice[u] = pos[r + 1] - pos[l];
else if (l == r) twice[u] = 0;
else if (cover[u] == 1) twice[u] = len[u << 1] + len[u << 1 | 1];
else twice[u] = twice[u << 1] + twice[u << 1 | 1];
}

void Modify(int u, int l, int r, int L, int R, int c) {
if (l == L && r == R) {
cover[u] += c;
getlen(u, l, r);
return ;
}
int mid = (l + r) >> 1;
if (R <= mid) Modify(u << 1, l, mid, L, R, c);
else if (L > mid) Modify(u << 1 | 1, mid + 1, r, L, R, c);
else {
Modify(u << 1, l, mid, L, mid, c);
Modify(u << 1 | 1, mid + 1, r, mid + 1, R, c);
}
getlen(u, l, r);
}

int find(double val) {
int l = 1, r = m;
while (l <= r) {
int mid = (l + r) >> 1;
if (pos[mid] == val) return mid;
else if (pos[mid] > val) r = mid - 1;
else l = mid + 1;
}
return -1;
}

void solve() {
double ans = 0;
for (int i = 1; i < cnt; i++) {
int l = find(S[i].l);
int r = find(S[i].r) - 1;
Modify(1, 1, m, l, r, S[i].f);
ans += (S[i + 1].h - S[i].h) * twice[1];
}
printf("%.2lf\n", ans);
}

int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}