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HDU 1828 Picture(矩形周长并)
HDU 1828 Picture
题目链接
题意:给定n个矩形,输出矩形周长并
思路:利用线段树去维护,分别从4个方向扫一次,每次多一段的时候,就查询该段未被覆盖的区间长度,然后周长就加上这个长度,4个方向都加完就是答案
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 5005; int n; struct Rec { int x1, y1, x2, y2; void read() { scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x1 += 10000; y1 += 10000; x2 += 10000; y2 += 10000; } } rec[N]; struct Line { int l, r, y, flag; Line() {} Line(int l, int r, int y, int flag) { this->l = l; this->r = r; this->y = y; this->flag = flag; } } line[N * 2]; bool cmp1(Line a, Line b) { if (a.y == b.y) return a.flag > b.flag; return a.y < b.y; } bool cmp2(Line a, Line b) { if (a.y == b.y) return a.flag > b.flag; return a.y > b.y; } #define lson(x) ((x<<1)+1) #define rson(x) ((x<<1)+2) struct Node { int l, r, add, addv, num; bool cover; void init(int l, int r) { this->l = l; this->r = r; add = addv = num = 0; cover = false; } void gao(int v) { addv += v; if (add == 0) num = r - l + 1; add += v; if (add == 0) num = 0; } } node[4 * 20005]; void pushup(int x) { if (node[lson(x)].cover && node[rson(x)].cover && node[lson(x)].add == node[rson(x)].add) { node[x].cover = true; node[x].add = node[lson(x)].add; } else node[x].cover = false; node[x].num = node[lson(x)].num + node[rson(x)].num; } void pushdown(int x) { if (node[x].addv) { node[lson(x)].gao(node[x].addv); node[rson(x)].gao(node[x].addv); node[x].addv = 0; } } void build(int l, int r, int x = 0) { node[x].init(l, r); if (l == r) { node[x].cover = true; return; } int mid = (l + r) / 2; build(l, mid, lson(x)); build(mid + 1, r, rson(x)); pushup(x); } void add(int l, int r, int v, int x = 0) { if (node[x].cover && node[x].l >= l && node[x].r <= r) { node[x].gao(v); return; } int mid = (node[x].l + node[x].r) / 2; pushdown(x); if (l <= mid) add(l, r, v, lson(x)); if (r > mid) add(l, r, v, rson(x)); pushup(x); } int query(int l, int r, int x = 0) { if (node[x].cover && node[x].l >= l && node[x].r <= r) return node[x].num; int mid = (node[x].l + node[x].r) / 2; pushdown(x); int ans = 0; if (l <= mid) ans += query(l, r, lson(x)); if (r > mid) ans += query(l, r, rson(x)); pushup(x); return ans; } int solve() { build(0, 20000); int ans = 0; for (int i = 0; i < 2 * n; i++) { if (line[i].flag) ans += line[i].r - line[i].l - query(line[i].l, line[i].r - 1); add(line[i].l, line[i].r - 1, line[i].flag); } return ans; } int gao() { int ans = 0; for (int i = 0; i < n; i++) { line[i * 2] = Line(rec[i].x1, rec[i].x2, rec[i].y1, 1); line[i * 2 + 1] = Line(rec[i].x1, rec[i].x2, rec[i].y2, -1); } sort(line, line + 2 * n, cmp1); ans += solve(); for (int i = 0; i < n; i++) { line[i * 2] = Line(rec[i].x1, rec[i].x2, rec[i].y2, 1); line[i * 2 + 1] = Line(rec[i].x1, rec[i].x2, rec[i].y1, -1); } sort(line, line + 2 * n, cmp2); ans += solve(); return ans; } int main() { while (~scanf("%d", &n)) { for (int i = 0; i < n; i++) rec[i].read(); int ans = 0; ans += gao(); for (int i = 0; i < n; i++) { swap(rec[i].x1, rec[i].y1); swap(rec[i].x2, rec[i].y2); } ans += gao(); printf("%d\n", ans); } return 0; }
HDU 1828 Picture(矩形周长并)
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