Poi 2001 goldmine line segment tree Scan Line

Question Link

Http://www.acm.cs.ecnu.edu.cn/problem.php? Problemid = 1350

Http://main.edu.pl/en/archive/oi/8/kop

Calculate N points (n <= 15000) in the plane, and use a matrix with a length and width of s w to set the maximum number of points. The points on the edge are also calculated.

In fact, it is similar to the previous rectangular nested area (poj's Atlantic). Using a scanning line like this, the vertex is treated as an edge (Y to Y value + W) and inserted into the line segment tree, in this way, the y direction is maintained, and the X direction is maintained using similar queues,

When the distance is greater than S, the Front (that is, the side removed from the line segment tree) is displayed, and the current side is added.

Maintain a value. It can be seen when the number of sides accumulates the most after scanning.

At first, I thought it was necessary to maintain the coverage value. Later I found it was useless. I just needed to maintain the accumulated layers directly.

At the beginning, I did not write the EOF, And re several times. I don't know why. After the change, I Will 1A.

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define lson RT <1, l, mid # define rson RT <1 | 1, Mid + 1, rusing namespace STD; const int n = 100020; // int cover [n <2]; int flag [n <2]; int d [n <2]; int S, W, N, maxn; struct node {int X, Y; bool operator <(const node & RHs) const {If (x = RHS. x) return Y <RHS. y; return x <RHS. X ;}} point [15010]; bool input () {maxn = 0; If (scanf ("% d", & S, & W) = EOF) return false; scanf ("% d", & N); For (INT I = 1; I <= N; I ++) {scanf ("% d ", & point [I]. x, & point [I]. y); point [I]. X ++ = 30000; point [I]. Y ++ = 30000; maxn = max (maxn, point [I]. y);} maxn + = W + 10; return true;} void build (int rt, int L, int R) {// cover [RT] = 0; flag [RT] = 0; d [RT] = 0; If (L> = r) return; int mid = (L + r)> 1; build (lson ); build (rson);} void Pushdown (int rt, int L, int R) {If (flag [RT] = 0) return; int mid = (L + r)> 1; // cover [RT <1] + = Flag [RT] * (mid-L + 1 ); // cover [RT <1 | 1] + = Flag [RT] * (R-mid); D [RT <1] + = Flag [RT]; d [RT <1 | 1] + = Flag [RT]; flag [RT <1] + = Flag [RT]; flag [RT <1 | 1] + = Flag [RT]; flag [RT] = 0;} void up (int rt) {// cover [RT] = cover [RT <1] + cover [RT <1 | 1]; d [RT] = max (d [RT <1], d [RT <1 | 1]);} void remove (int l, int R, int RT, int L, int R) {If (L <= L & R <= r) {// cover [RT]-= (R-l + 1 ); d [RT] --; flag [RT] + =-1; return;} Pushdown (RT, L, R); int mid = (L + r)> 1; if (L <= mid) Remove (L, R, lson); If (r> mid) Remove (L, R, rson); up (RT );} void inserts (int l, int R, int RT, int L, int R) {If (L <= L & R <= r) {// cover [RT] + = (R-l + 1); D [RT] ++; flag [RT] + = 1; return;} Pushdown (RT, l, R); int mid = (L + r)> 1; if (L <= mid) inserts (L, R, lson); If (r> mid) inserts (L, R, rson); up (RT) ;}int main () {While (input () {sort (point + 1, point + 1 + n ); build (1, 0, maxn); int pre = 1; int ans = 0; For (INT I = 1; I <= N; I ++) {// cout <I <Endl; while (point [pre]. X + S <point [I]. x) {remove (point [pre]. y, point [pre]. Y + W, 1, 0, maxn); Pre ++;} inserts (point [I]. y, point [I]. Y + W, 1, 0, maxn); ans = max (ANS, d [1]);} printf ("% d \ n", ANS );}}

Poi 2001 goldmine line segment tree Scan Line