扫描线 hdu1542
扫描线分类两种,一种离散y轴,一种离散x轴。以离垂直x轴的直线为例:
将每个离散的线之间作为一个线段树的叶子节点,然后从最小与x轴平行的直线开始,区覆盖线段树的叶子节点,
也就是两根与y轴平行的直线之间所夹的空间,这段空间的长度乘以你的扫描线之间的距离(与x轴平行的直线),就是面积,
看图吧,图下的数字是代表覆盖的次数,矩形的下底线扫描覆盖节点时为 value +=1,上边线扫描 value -= 1;
代码如下
#include <map> #include <cstring> #include <cstdio> #include <algorithm> #include <map> using namespace std; const int maxn = 205; struct node { int r, l, lazy; double dat; }tr[maxn<<2]; struct bian { double x, a, b; int k; }egde[maxn<<1]; int n; double yy[maxn << 1]; map<double, int> mp; //注意节点为 l+1=r,离散化会丢失离散节点之间的信息 void build(int p, int l, int r) { tr[p].dat = 0; tr[p].l = l;tr[p].r = r; tr[p].lazy = 0; if (l == r - 1)return; int mid = (l + r) >> 1; build(p << 1, l, mid); build(p << 1 | 1, mid, r); } inline void push(int p) { if (tr[p].lazy)tr[p].dat = yy[tr[p].r] - yy[tr[p].l]; else if (tr[p].l + 1 == tr[p].r)tr[p].dat = 0; else tr[p].dat = tr[p << 1].dat + tr[p << 1 | 1].dat; } void updata(int p, int l, int r, int k) { if (tr[p].l > r || tr[p].r < l) return; if (l <= tr[p].l && tr[p].r <= r) { tr[p].lazy += k; push(p); return; } int mid = (tr[p].l + tr[p].r) >> 1; if (l <= mid) updata(p << 1, l, r, k); if (mid < r) updata(p << 1 | 1, l, r, k); push(p); } bool cmp(const bian& a, const bian& b) { if(a.x != b.x)return a.x < b.x; return a.k > b.k; } int main(void) { /*freopen("atlantis.in", "r", stdin); freopen("atlantis.out", "w", stdout);*/ for (int cas = 1;~scanf("%d", &n)&&n; ++cas) { double ans = 0; for (int i = 0; i < n; ++i) { double x, y, a, b; scanf("%lf%lf%lf%lf", &x, &y, &a, &b); egde[i << 1] = { x, y, b, 1 }; egde[i << 1 | 1] = { a, y, b, -1 }; yy[i << 1] = y;yy[i << 1 | 1] = b; } sort(yy, yy + (n << 1));sort(egde, egde + (n << 1), cmp); int tot = unique(yy, yy + (n << 1)) - yy; for (int i = 0; i < tot; ++i)mp[yy[i]] = i; build(1, 0, tot - 1); updata(1, mp[egde[0].a], mp[egde[0].b], egde[0].k); for (int i = 1; i < (n << 1); ++i) { ans += (egde[i].x - egde[i - 1].x) * tr[1].dat; updata(1, mp[egde[i].a], mp[egde[i].b], egde[i].k); } printf("Test case #%d\nTotal explored area: %.2lf\n\n", cas, ans); } return 0; }