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。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;
}