poj 2482 Stars in Your Window(扫描线)

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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。只要以每个点为左上角，建立矩形，这个矩形即为框框左下角放的位置可以括到该点，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;}

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