poj1177
题意：在平面直角坐标系内给出一些与坐标轴平行的矩形，将这些矩形覆盖的区域求并集，然后问这个区域的周长是多少。（边与边重合的地方不计入周长）
分析：线段树。曾经做过类似的求矩形覆盖的总面积的题。这道题同样要使用扫描线算法。属于线保留型线段树。
我们先领扫描线与y轴平行。
线段树内每个节点除了要记录该区间被覆盖了几层之外，还要记录当前状态下扫描线在该区间（开区间）内与多少条与x轴平行的边相交。
节点上还有两个bool型变量，记录该区间内（包括子树）线段覆盖是否接触到该区间的起始和结束点。
在父节点如果没被整个覆盖，则需要从子区间的起始和结束点来更新父节点两端点的覆盖情况。
更新过程在返回时，父节点的交点数量应等于两子节点交点数量的和，另外特判一下两子区间的公共点（父节点的中点），判断这里是不是覆盖与未覆盖的分界点，如果是则还需要在父节点上增加这个交点。
这样就得知了每段与x轴平行的距离内有多少条线段需要计算，距离乘以数量即可。这样就计算出了所有与x轴平行的周长上的边的总长度。
之后让扫描线与x轴平行即可计算出与y轴平行的所有周长上的边的总长度。
#include <cstdio> #include <algorithm> using namespace std; //scanning from left to right //discretionize Ys #define MAX_REC_NUM 5005 #define MAX_INTERVAL MAX_REC_NUM * 2 struct Node { int l, r; Node *pleft, *pright; int num; bool to_left, to_right; int edge_num; }; int node_cnt; Node tree[MAX_INTERVAL * 3]; struct Interval { int start, end; int pos; int value; Interval() {} Interval(int start, int end, int pos, int value):start(start), end(end), pos(pos), value(value) {} bool operator < (const Interval &a)const { if (pos != a.pos) return pos < a.pos; return value > a.value; } }interval[MAX_REC_NUM * 2]; struct Rectangle { int l, d, u, r; }rec[MAX_REC_NUM]; int discrete[MAX_REC_NUM * 2]; int discrete_num; int rec_num; int interval_num; int get_index(int a) { return lower_bound(discrete, discrete + discrete_num, a) - discrete; } void discretization(int discrete[], int &discrete_num) { sort(discrete, discrete + discrete_num); discrete_num = unique(discrete, discrete + discrete_num) - discrete; } void input() { scanf("%d", &rec_num); for (int i = 0; i < rec_num; i++) { int l, d, r, u; scanf("%d%d%d%d", &l, &d, &r, &u); rec[i].l = l; rec[i].r = r; rec[i].u = u; rec[i].d = d; } } void make_xscan() { discrete_num = 0; interval_num = 0; for (int i = 0; i < rec_num; i++) { int l, d, r, u; l = rec[i].l; r = rec[i].r; u = rec[i].u; d = rec[i].d; interval[interval_num++] = Interval(d, u, l, 1); interval[interval_num++] = Interval(d, u, r, -1); discrete[discrete_num++] = u; discrete[discrete_num++] = d; } } void make_yscan() { discrete_num = 0; interval_num = 0; for (int i = 0; i < rec_num; i++) { int l, d, r, u; l = rec[i].l; r = rec[i].r; u = rec[i].u; d = rec[i].d; interval[interval_num++] = Interval(l, r, d, 1); interval[interval_num++] = Interval(l, r, u, -1); discrete[discrete_num++] = l; discrete[discrete_num++] = r; } } void buildtree(Node *proot, int s, int e) { proot->l = s; proot->r = e; proot->to_left = false; proot->to_right = false; proot->num = 0; proot->edge_num = 0; if (s == e) { proot->pleft = proot->pright = NULL; return; } node_cnt++; proot->pleft = tree + node_cnt; node_cnt++; proot->pright = tree + node_cnt; buildtree(proot->pleft, s, (s + e) / 2); buildtree(proot->pright, (s + e) / 2 + 1, e); } void recount(Node *p) { if (p->num > 0) { p->edge_num = 0; p->to_right = p->to_left = true; return; } if (p->pleft == NULL || p->pright == NULL) { p->edge_num = 0; p->to_right = p->to_left = false; return; } p->to_left = p->pleft->to_left; p->to_right = p->pright->to_right; p->edge_num = p->pleft->edge_num + p->pright->edge_num; if (p->pleft->to_right != p->pright->to_left) p->edge_num++; } void insert(Node *proot, int s, int e, int value) { if (s > proot->r || e < proot->l) return; s = max(s, proot->l); e = min(e, proot->r); if (s == proot->l && e == proot->r) { proot->num += value; recount(proot); return; } insert(proot->pleft, s, e, value); insert(proot->pright, s, e, value); recount(proot); } long long work() { long long ans = 0; for (int i = 0; i < interval_num; i++) { int s = get_index(interval[i].start); int e = get_index(interval[i].end) - 1; insert(tree, s, e, interval[i].value); long long line_num = tree->edge_num; if (tree->to_left) line_num++; if (tree->to_right) line_num++; if (i != interval_num - 1) ans += (interval[i + 1].pos - interval[i].pos) * line_num; } return ans; } int main() { input(); long long ans = 0; make_xscan(); sort(interval, interval + interval_num); discretization(discrete, discrete_num); buildtree(tree, 0, discrete_num); ans += work(); make_yscan(); sort(interval, interval + interval_num); discretization(discrete, discrete_num); buildtree(tree, 0, discrete_num); ans += work(); printf("%lld\n", ans); return 0; }