题目链接：hdu 5091 Beam Cannon

题目大意：给定N个点，如今要有一个W∗H的矩形，问说最多能圈住多少个点。

解题思路：线段的扫描线，如果有点(x,y)，那么(x,y)~(x+W,y+H)形成的矩形，以框的右下角落的位置是能够圈住(x,y)

点，所以N个点即为N个矩形，求覆盖的最大次数，扫描线裸题。

#include <cstdio>

#include <cstring>

#include <vector>

#include <algorithm>

using namespace std;

const int maxn = 40005;

const int bw = 20000;

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

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

int lc[maxn << 2], rc[maxn << 2], ad[maxn << 2], nd[maxn << 2];

inline void pushup(int u) {

    nd[u] = max(nd[lson(u)], + nd[rson(u)]) + ad[u];

}

inline void maintain(int u, int v) {

    ad[u] += v;

    nd[u] += v;

}

void build(int u, int l, int r) {

    lc[u] = l;

    rc[u] = r;

    ad[u] = nd[u] = 0;

    if (l == r)

        return;

    int mid = (l + r) >> 1;

    build(lson(u), l, mid);

    build(rson(u), mid + 1, r);

    pushup(u);

}

void modify(int u, int l, int r, int v) {

    if (l <= lc[u] && rc[u] <= r) {

        maintain(u, v);

        return;

    }

    int mid = (lc[u] + rc[u]) >> 1;

    if (l <= mid)

        modify(lson(u), l, r, v);

    if (r > mid)

        modify(rson(u), l, r, v);

    pushup(u);

}

int query(int u, int l, int r) {

    if (l <= lc[u] && rc[u] <= r)

        return nd[u];

    int mid = (lc[u] + rc[u]) >> 1, ret = 0;

    if (l <= mid)

        ret = max(ret, query(lson(u), l, r));

    if (r > mid)

        ret = max(ret, query(rson(u), l, r));

    pushup(u);

    return ret;

}

struct Seg {

    int x, l, r, w;

    Seg(int x = 0, int l = 0, int r = 0, int w = 0) {

        this->x = x;

        this->l = l;

        this->r = r;

        this->w = w;

    }

};

vector<Seg> G;

inline bool cmp (const Seg& a, const Seg& b) {

    if (a.x != b.x)

        return a.x < b.x;

    return a.w < b.w;

}

int N, W, H;

void init () {

    int x, y;

    G.clear();

    scanf("%d%d", &W, &H);

    for (int i = 0; i < N; i++) {

        scanf("%d%d", &x, &y);

        G.push_back(Seg(x, y + bw, min(bw, y + H) + bw, 1));

        G.push_back(Seg(x + W + 1, y + bw, min(bw, y + H) + bw, -1));

    }

    build(1, 0, bw * 2);

    sort(G.begin(), G.end(), cmp);

}

int main () {

    while (scanf("%d", &N) == 1 && N != -1) {

        init();

        int ans = 0;

        for (int i = 0; i < G.size(); i++) {

            modify(1, G[i].l, G[i].r, G[i].w);

            ans = max(ans, nd[1]);

        }

        printf("%d\n", ans);

    }

    return 0;

}