poj 2482 Stars in Your Window(扫描线)

题目链接：poj 2482 Stars in Your Window

题目大意：平面上有N个星星。问一个W∗H的矩形最多能括进多少个星星。

解题思路：扫描线的变形。仅仅要以每一个点为左上角。建立矩形，这个矩形即为框框左下角放的位置能够括到该点，那么N个星星就有N个矩形，扫描线处理哪些位置覆盖次数最多。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;
const int maxn = 40000;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)

int lc[maxn << 2], rc[maxn << 2];
ll v[maxn << 2], s[maxn << 2];

inline void pushup (int u) {
s[u] = max(s[lson(u)], s[rson(u)]) + v[u];
}

inline void maintain (int u, int d) {
v[u] += d;
pushup(u);
}

void build (int u, int l, int r) {
lc[u] = l;
rc[u] = r;
v[u] = s[u] = 0;

if (l == r)
return;

int mid = (l + r) / 2;
build(lson(u), l, mid);
build(rson(u), mid + 1, r);
pushup(u);
}

void modify (int u, int l, int r, int d) {
if (l <= lc[u] && rc[u] <= r) {
maintain(u, d);
return;
}

int mid = (lc[u] + rc[u]) / 2;
if (l <= mid)
modify(lson(u), l, r, d);
if (r > mid)
modify(rson(u), l, r, d);
pushup(u);
}

struct Seg {
ll x, l, r, d;
Seg (ll x = 0, ll l = 0, ll r = 0, ll d = 0) {
this->x = x;
this->l = l;
this->r = r;
this->d = d;
}
friend bool operator < (const Seg& a, const Seg& b) {
return a.x < b.x;
}
};

int N, W, H;
vector<ll> pos;
vector<Seg> vec;

inline int find (ll k) {
return lower_bound(pos.begin(), pos.end(), k) - pos.begin();
}

void init () {
ll x, y, d;

pos.clear();
vec.clear();
for (int i = 0; i < N; i++) {
scanf("%lld%lld%lld", &x, &y, &d);
pos.push_back(y - H);
pos.push_back(y);
vec.push_back(Seg(x - W, y - H, y, d));
vec.push_back(Seg(x, y - H, y, -d));
}
sort(pos.begin(), pos.end());
sort(vec.begin(), vec.end());
}

ll solve () {
ll ret = 0;
build (1, 0, pos.size());

for (int i = 0; i < vec.size(); i++) {
modify(1, find(vec[i].l), find(vec[i].r) - 1, vec[i].d);
ret = max(ret, s[1]);
}
return ret;
}

int main () {
while (scanf("%d%d%d", &N, &W, &H) == 3) {
init();
printf("%lld\n", solve());
}
return 0;
}