hdu 5091 Beam Cannon(扫描线段树)

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;
}

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