POJ 2482 Stars in Your Window(线段树扫描线)

题意:

给出星星的坐标和亮度,给定一个矩形(长/宽),求框住的星星亮度和的最大值,恰好在边上的不算。

对于每颗星星，我们以左下角构成一个矩形，两条边：一条范围[x,x+w),高度为y,权值为brightness，另一条范围[x,x+w)，高度为y+h,亮度为-brightness。这样我们对所有边进行排序，从低到高扫描就行了。计算线段最大覆盖次数即可。

#include <cstdio>#include <cstring>#include <cctype>#include <algorithm>using namespace std;#define MID(x,y) ( ( x + y ) >> 1 )  #define L(x) ( x << 1 )  #define R(x) ( x << 1 | 1 )  #define FOR(i,s,t) for(int i=s; i<t; i++)  using namespace std;const int MAX = 20010;struct point { int x, y; long long lx, ly, rx, ry; int val; };point p[MAX / 2];long long x[MAX], y[MAX];int ans;struct Sline { int x, y1, y2; int flag; };Sline l[MAX];void add_line(int x1, int y1, int x2, int y2, int &cnt, int flag){l[cnt].x = x1; l[cnt].y1 = y1; l[cnt].y2 = y2;l[cnt++].flag = flag;l[cnt].x = x2; l[cnt].y1 = y1; l[cnt].y2 = y2;l[cnt++].flag = -flag;}bool cmp(Sline a, Sline b){if (a.x == b.x) return a.flag < b.flag;return a.x < b.x;}struct Tnode {               // 一维线段树   int l, r, val, max;int len() { return r - l; }int mid() { return MID(l, r); }bool in(int ll, int rr) { return l >= ll && r <= rr; }void lr(int ll, int rr) { l = ll; r = rr; }};Tnode node[MAX << 2];void Build(int t, int l, int r){node[t].val = node[t].max = 0;node[t].lr(l, r);if (node[t].len() == 1) return;int mid = MID(l, r);Build(L(t), l, mid);Build(R(t), mid, r);}void Updata(int t, int l, int r, int val){if (node[t].in(l, r)){node[t].val += val;node[t].max += val;ans = max(node[t].max, ans);return;}if (node[t].len() == 1) return;node[L(t)].val += node[t].val;node[L(t)].max += node[t].val;node[R(t)].val += node[t].val;node[R(t)].max += node[t].val;node[t].val = 0;int mid = node[t].mid();if (l >= mid)Updata(R(t), l, r, val);elseif (r <= mid)Updata(L(t), l, r, val);else{Updata(L(t), l, r, val);Updata(R(t), l, r, val);}node[t].max = max(node[L(t)].max, node[R(t)].max);ans = max(node[t].max, ans);}int solve(int cx, int cy, int n){sort(x, x + cx);sort(y, y + cy);cx = unique(x, x + cx) - x;cy = unique(y, y + cy) - y;int cnt = 0;FOR(i, 0, n){int x2 = lower_bound(x, x + cx, p[i].rx) - x;int y2 = lower_bound(y, y + cy, p[i].ry) - y;int x1 = lower_bound(x, x + cx, p[i].lx) - x;int y1 = lower_bound(y, y + cy, p[i].ly) - y;add_line(x1, y1, x2, y2, cnt, p[i].val);}sort(l, l + cnt, cmp);Build(1, 0, cy + 1);FOR(i, 0, cnt)Updata(1, l[i].y1, l[i].y2, l[i].flag);return ans;}int main(){int l;int n, w;while (scanf("%d%d%d", &n, &w, &l) != EOF){int cntx = 0, cnty = 0;ans = 0;FOR(i, 0, n){scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].val);p[i].lx = p[i].x; p[i].ly = p[i].y;p[i].rx = p[i].x * 1ll + w; p[i].ry = p[i].y * 1ll + l;x[cntx++] = p[i].lx;x[cntx++] = p[i].rx;y[cnty++] = p[i].ly;y[cnty++] = p[i].ry;}solve(cntx, cnty, n);printf("%d\n", ans);}return 0;}