题意:给你一些矩形的左上角点的坐标和右下角点的坐标,求周长并
最显而易见的思路就是对于x轴和y轴做两次扫描线,对于负数的坐标进行离散化。每次增加的值是线段变化量的绝对值。具体写法和求面积并的差不多。
#include <cstdio> #include <algorithm> #include <cstring> #include <vector> using namespace std; #define lson rt << 1 , l , mid #define rson rt << 1 | 1, mid + 1 , r const int maxn = 10000 + 10; int cnt[maxn << 2], len[maxn << 2], n; struct Seg { int x, l, r, cover; Seg(int x, int l, int r, int cover) :x(x), l(l), r(r), cover(cover) {} bool operator < (const Seg &seg) const { if (x == seg.x) return cover > seg.cover; return x < seg.x; } }; vector<Seg> seg; vector<int> num; int sx1[maxn], sy1[maxn], sx2[maxn], sy2[maxn]; void pushup(int rt, int l, int r) { if (cnt[rt]) len[rt] = num[r + 1] - num[l]; else if (l == r) len[rt] = 0; else len[rt] = len[rt << 1] + len[rt << 1 | 1]; } void update(int rt, int l, int r, int ql, int qr, int Val) { if (ql <= l && qr >= r) { cnt[rt] += Val; pushup(rt, l, r); } else { int mid = (l + r) >> 1; if (ql <= mid) update(lson, ql, qr, Val); if (qr > mid) update(rson, ql, qr, Val); pushup(rt, l, r); } } int GetID(int val) { return lower_bound(num.begin(), num.end(), val) - num.begin(); } int solve(int px1[], int py1[], int px2[], int py2[]) { seg.clear(); num.clear(); memset(cnt, 0, sizeof(cnt)); memset(len, 0, sizeof(len)); for (int i = 0; i < n; i++) { seg.push_back(Seg(px1[i], py1[i], py2[i], 1)); seg.push_back(Seg(px2[i], py1[i], py2[i], -1)); num.push_back(py1[i]); num.push_back(py2[i]); } sort(seg.begin(), seg.end()); sort(num.begin(), num.end()); num.erase(unique(num.begin(), num.end()), num.end()); int ret = 0, msize = seg.size(), nowlen = 0, k = num.size(); for (int i = 0; i < msize; i++) { int ql = GetID(seg[i].l), qr = GetID(seg[i].r) - 1; update(1, 0, k - 1, ql, qr, seg[i].cover); ret += abs(nowlen - len[1]); nowlen = len[1]; } return ret; } int main() { while (scanf("%d", &n) != EOF) { for (int i = 0; i < n; i++) { scanf("%d%d%d%d", &sx1[i], &sy1[i], &sx2[i], &sy2[i]); } int ret = solve(sx1, sy1, sx2, sy2) + solve(sy1, sx1, sy2, sx2); printf("%d\n", ret); } }
总觉得可以就做一次扫描线就可以完成统计,就是不知道应该怎么写。看了别人的blog之后发现自己还是太弱。
要统计的东西有这些,假设扫描的是y轴。
1. y轴上有多少条线段,并且如果端点重合的话就算一条。因为x轴上面有多少条线是和y轴上有多少条线段有关的,是2倍的关系。为此还要开两个数组来辅助标记,表示当前线段的左端点和右端点是否被覆盖。
2. y轴上点段的长度,这个和上面的做法统计的功能一样。
3. 当前线段被覆盖了多少次,这个也和上面的统计方法一样。
#include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <climits> using namespace std; const int maxn = 2e4 + 10; #define lson rt << 1, l, mid #define rson rt << 1 | 1,mid + 1,r struct Seg { int l, r, x, cover; Seg(int l, int r, int x, int cover) :l(l), r(r), x(x), cover(cover) {} bool operator < (const Seg &seg) const { if (x == seg.x) return cover > seg.cover; return x < seg.x; } }; vector<Seg> seg; int n; bool lbound[maxn << 2], rbound[maxn << 2]; int numseg[maxn << 2], len[maxn << 2], cnt[maxn << 2]; int minval, maxval; void pushup(int rt, int l, int r) { int lc = rt << 1, rc = rt << 1 | 1; if (cnt[rt]) { lbound[rt] = rbound[rt] = true; //因为每一条线段的长度都等于1,第l条线段起始点是l,第r条线段的结束点是r + 1,长度是r + 1 - l len[rt] = r - l + 1; //因为横向必定有两条边,所以为2 numseg[rt] = 2; } else if (l == r) { len[rt] = numseg[rt] = lbound[rt] = rbound[rt] = 0; } else { len[rt] = len[lc] + len[rc]; lbound[rt] = lbound[lc]; rbound[rt] = rbound[rc]; numseg[rt] = numseg[lc] + numseg[rc]; //如果在交点重合 if (rbound[lc] && lbound[rc]) { numseg[rt] -= 2; } } } void update(int rt, int l, int r, int ql, int qr, int Val) { if (ql <= l && qr >= r) { cnt[rt] += Val; pushup(rt, l, r); } else { int mid = (l + r) >> 1; if (ql <= mid) update(lson, ql, qr, Val); if (qr > mid) update(rson, ql, qr, Val); pushup(rt, l, r); } } int main() { while (scanf("%d", &n) != EOF) { seg.clear(); minval = INT_MAX; maxval = -1; for (int i = 1; i <= n; i++) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); seg.push_back(Seg(y1, y2, x1, 1)); seg.push_back(Seg(y1, y2, x2, -1)); minval = min(minval, min(y1, y2)); maxval = max(maxval, max(y1, y2)); } sort(seg.begin(), seg.end()); int msize = seg.size(), ret = 0, last = 0; for (int i = 0; i < msize; i++) { if (seg[i].l < seg[i].r) update(1, minval, maxval, seg[i].l, seg[i].r - 1, seg[i].cover); //横向的长度 if (i < msize - 1) ret += numseg[1] * (seg[i + 1].x - seg[i].x); //纵向的长度 ret += abs(last - len[1]); last = len[1]; } printf("%d\n", ret); } return 0; }
HDU 1828 Picture 线段树+扫描线,布布扣,bubuko.com
原文:http://www.cnblogs.com/rolight/p/3917645.html