119 lines
3.6 KiB
Markdown
119 lines
3.6 KiB
Markdown
##[$HDU5091$ $Beam$ $Cannon$](http://acm.hdu.edu.cn/showproblem.php?pid=5091)
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### 一、题目大意
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有$n$个点($n<=10000$)，点的坐标绝对值不超过$20000$，然后问你用一个$w*h(1<=w,h<=40000)$的矩形，矩形的边平行于坐标轴，最多能盖住多少个点。
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刘汝佳黑书上有原题
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<center><img src='https://images0.cnblogs.com/blog2015/710860/201504/302330388653886.png'></center>
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<center><img src='https://images0.cnblogs.com/blog2015/710860/201504/302330537712033.png'></center>
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<center><img src='https://images0.cnblogs.com/blog2015/710860/201504/302331078335136.png'></center>
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下面这份代码是加了离散化的，用垂直于$x$轴的直线去扫描，在$y$轴上建立线段树，所以对于每一个点$(x,y)$，赋予权值$1$,，然后增加一个新的负点$(x+w,y)$，赋予权值$-1$。当扫描线扫过每一个点，用该点的权值去区间更新线段树。维护每个区间的$sum$值，因为是区间更新，所以还要打个$lazy$标记。
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### 二、实现代码
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```cpp {.line-numbers}
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#include <cstdio>
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#include <cstring>
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#include <algorithm>
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#include <iostream>
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#include <vector>
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using namespace std;
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const int N = 40010;
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struct Node {
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int l, r, lazy;
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int maxc;
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} tr[N << 2];
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struct Seg {
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int x, y1, y2, c;
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bool operator<(const Seg &t) const {
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if (x == t.x) return c > t.c;
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return x < t.x;
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}
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} seg[N << 1];
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void pushup(int u) {
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tr[u].maxc = max(tr[u << 1].maxc, tr[u << 1 | 1].maxc);
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}
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void pushdown(int u) {
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if (tr[u].lazy) {
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tr[u << 1].lazy += tr[u].lazy;
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tr[u << 1 | 1].lazy += tr[u].lazy;
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tr[u << 1].maxc += tr[u].lazy;
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tr[u << 1 | 1].maxc += tr[u].lazy;
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tr[u].lazy = 0;
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}
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}
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void build(int u, int l, int r) {
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tr[u] = {l, r, 0, 0};
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if (l == r) return;
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int mid = (l + r) >> 1;
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build(u << 1, l, mid);
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build(u << 1 | 1, mid + 1, r);
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}
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void modify(int u, int l, int r, int v) {
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if (l <= tr[u].l && tr[u].r <= r) {
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tr[u].lazy += v;
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tr[u].maxc += v;
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return;
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}
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pushdown(u);
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int mid = tr[u].l + tr[u].r >> 1;
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if (l <= mid) modify(u << 1, l, r, v);
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if (r > mid) modify(u << 1 | 1, l, r, v);
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pushup(u);
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}
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//离散化数组
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vector<int> ys;
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int find(int x) {
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return lower_bound(ys.begin(), ys.end(), x) - ys.begin();
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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int n, w, h;
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while (cin >> n && ~n) {
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//多组数测数据
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memset(tr, 0, sizeof tr);
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ys.clear();
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int idx = 0;
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cin >> w >> h;
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for (int i = 1; i <= n; i++) {
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int x, y;
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cin >> x >> y;
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seg[++idx] = {x, y, y + h, 1};
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seg[++idx] = {x + w, y, y + h, -1};
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ys.push_back(y);
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ys.push_back(y + h);
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}
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//对于边，按x由小到大，入边在前，出边在后排序
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sort(seg + 1, seg + 1 + idx);
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//离散化=排序+ 去重
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sort(ys.begin(), ys.end());
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ys.erase(unique(ys.begin(), ys.end()), ys.end());
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//离散化似乎应该成为必要的步骤，这样记忆起来省的麻烦，都是一样的套路才方便记忆
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build(1, 0, ys.size() - 1); // 0~ys.size()-1,共 ys.size()个,线段树是建立在y轴的投影上
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int ans = 0;
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for (int i = 1; i <= idx; i++) {
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modify(1, find(seg[i].y1), find(seg[i].y2), seg[i].c);
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ans = max(ans, tr[1].maxc);
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}
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printf("%d\n", ans);
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}
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return 0;
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}
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``` |