题面据说很美～
每个星星可以根据在窗口的左下角和右上角两个位置建立两条扫描线，之后就是简单的区间增减和求最大值操作了。
注意要处理在边界上的星星是不算的情况，其实只要把左右边界分别增减一个eps即可。
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define MP make_pair #define PB push_back #define lson rt <<1,l,mid #define rson rt <<1 | 1,mid + 1,r typedef long long LL; typedef unsigned long long ULL; typedef vector VI; typedef pair pii; const int INF = INT_MAX / 3; const double eps = 1e-4; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 1e5 + 5; struct Seg { double x,l,r; int cover; Seg(double x,double l,double r,int cover): x(x),l(l),r(r),cover(cover) {} bool operator <(const Seg &s) const { return x s; vector numy; LL lazy[maxn <<2],maxv[maxn <<2]; int W,H,n; void add_line(int x,int y,int cover) { double x1 = x - W + eps, x2 = x - eps, y1 = y - H + eps, y2 = y - eps; s.PB(Seg(x1,y1,y2,cover)); s.PB(Seg(x2,y1,y2,-cover)); numy.PB(y1); numy.PB(y2); } int getID(double Val) { return lower_bound(numy.begin(),numy.end(),Val) - numy.begin(); } void pushdown(int rt,int l,int r) { if(lazy[rt] == 0) return; int lc = rt <<1,rc = rt <<1 | 1; lazy[lc] += lazy[rt]; lazy[rc] += lazy[rt]; maxv[lc] += lazy[rt]; maxv[rc] += lazy[rt]; lazy[rt] = 0; } void pushup(int rt,int l,int r) { maxv[rt] = max(maxv[rt <<1] ,maxv[rt <<1 | 1]); } void update(int rt,int l,int r,int ql,int qr,int Val) { if(ql <= l && qr >= r) { lazy[rt] += Val; maxv[rt] += Val; } else { pushdown(rt,l,r); 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); } } void solve() { sort(numy.begin(),numy.end()); sort(s.begin(),s.end()); int ks = s.size(), ky = numy.size(); LL ans = 0; for(int i = 0;i
POJ 2482 Stars in Your Window 线段树+离散化+扫描线,,
POJ 2482 Stars in Your Window 线段树+离散化+扫描线
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