Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

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用于快速求矩形面积并。

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#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
typedef long long LL;
int n, len, d[N << 1];
int get(int x)
{
return lower_bound(d + 1, d + 1 + len, x) - d;
}
struct Line
{
int x, y1, y2, op;
bool operator<(const Line &_) const
{
return x < _.x;
}
} e[N << 1];
struct Node
{
int l, r, cnt, len;
} tr[N << 3];
void build(int x, int l, int r)
{
tr[x].l = l, tr[x].r = r;
if (l == r)
return;
int mid = (l + r) >> 1;
build(x << 1, l, mid);
build(x << 1 | 1, mid + 1, r);
}
void modify(int x, int l, int r, int k)
{
if (tr[x].l >= l && tr[x].r <= r)
{
tr[x].cnt += k;
if (tr[x].cnt > 0)
tr[x].len = d[tr[x].r + 1] - d[tr[x].l];
else
tr[x].len = (tr[x].l == tr[x].r ? 0 : tr[x << 1].len + tr[x << 1 | 1].len);
return;
}
int mid = (tr[x].l + tr[x].r) >> 1;
if (l <= mid)
modify(x << 1, l, r, k);
if (r > mid)
modify(x << 1 | 1, l, r, k);
tr[x].len = (tr[x].cnt > 0 ? d[tr[x].r + 1] - d[tr[x].l] : tr[x << 1].len + tr[x << 1 | 1].len);
}
int main()
{
scanf("%d", &n);
len = 0;
for (int i = 1; i <= n; i++)
{
int x1, y1, x2, y2;
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
d[++len] = y1, d[++len] = y2;
e[i] = (Line){x1, y1, y2, 1};
e[i + n] = (Line){x2, y1, y2, -1};
}
sort(d + 1, d + len + 1);
len = unique(d + 1, d + len + 1) - (d + 1);
sort(e + 1, e + 2 * n + 1);
build(1, 1, len);
LL ans = 0;
for (int i = 1; i <= 2 * n; i++)
{
modify(1, get(e[i].y1), get(e[i].y2) - 1, e[i].op);
if (i < 2 * n && e[i].x != e[i + 1].x)
ans += (LL)(e[i + 1].x - e[i].x) * tr[1].len;
}
printf("%lld", ans);
return 0;
}