2018-2019 ACM-ICPC, Asia Seoul Regional Contest

ProblemA Circuits

Solved.

题意：

有$n$个矩形，可以放两条平行与$x$轴的线，求怎么放置两条无线长的平行于$x$轴的线，使得他们与矩形相交个数最多

如果一个矩形同时与两条线相交，只算一次。

思路：

离散化后枚举一根线，另一根线用线段树维护，扫描线思想

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 #define N 400010
  5 int n;
  6 int b[N]; 
  7 int x[N], y[N];
  8 vector <int> in[N], out[N];
  9 int ans[N];
 10 
 11 namespace SEG
 12 {
 13     struct node
 14     {
 15         int Max, lazy;
 16         node () {}
 17         node (int Max, int lazy) : Max(Max), lazy(lazy) {}
 18         void init() { Max = lazy = 0; }
 19         void add(int x)
 20         {
 21             Max += x;
 22             lazy += x;
 23         }
 24         node operator + (const node &other) const
 25         {
 26             node res; res.init();
 27             res.Max = max(Max, other.Max);
 28             return res;
 29         }
 30     }a[N << 2];
 31     void build(int id, int l, int r)
 32     {
 33         a[id].init();
 34         if (l == r)
 35             return;
 36         int mid = (l + r) >> 1;
 37         build(id << 1, l, mid);
 38         build(id << 1 | 1, mid + 1, r);
 39     }
 40     void pushdown(int id)
 41     {
 42         if (!a[id].lazy) return;
 43         a[id << 1].add(a[id].lazy);
 44         a[id << 1 | 1].add(a[id].lazy);
 45         a[id].lazy = 0;
 46     }
 47     void update(int id, int l, int r, int ql, int qr, int x)
 48     {
 49         if (l >= ql && r <= qr)
 50         {
 51             a[id].add(x);
 52             return;
 53         }
 54         int mid = (l + r) >> 1;
 55         pushdown(id);
 56         if (ql <= mid) update(id << 1, l, mid, ql, qr, x);
 57         if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr, x);
 58         a[id] = a[id << 1] + a[id << 1 | 1];
 59     } 
 60     int query(int id, int l, int r, int pos)
 61     {
 62         if (l == r) return a[id].Max;
 63         int mid = (l + r) >> 1;
 64         pushdown(id);
 65         if (pos <= mid) return query(id << 1, l, mid, pos);
 66         else return query(id << 1 | 1, mid + 1, r, pos);
 67     }
 68 }
 69 
 70 void Hash()
 71 {
 72     sort(b + 1, b + 1 + b[0]);
 73     b[0] = unique(b + 1, b + 1 + b[0]) - b - 1;
 74     for (int i = 1; i <= n; ++i) x[i] = lower_bound(b + 1, b + 1 + b[0], x[i]) - b;
 75     for (int i = 1; i <= n; ++i) y[i] = lower_bound(b + 1, b + 1 + b[0], y[i]) - b;
 76 }
 77 
 78 int main()
 79 {
 80     while (scanf("%d", &n) != EOF)
 81     {
 82         b[0] = 0;  
 83         for (int i = 1; i < N; ++i)
 84             in[i].clear(), out[i].clear();
 85         for (int i = 1, tmp; i <= n; ++i)
 86         {
 87             scanf("%d%d%d%d", &tmp, y + i, &tmp, x + i);
 88 //            cout << x[i] << " " << y[i] << endl;
 89             b[++b[0]] = x[i];
 90             b[++b[0]] = y[i];
 91         }
 92         Hash();
 93         SEG::build(1, 1, b[0]);
 94         for (int i = 1; i <= n; ++i)
 95         {
 96             in[x[i]].push_back(i);
 97             out[y[i]].push_back(i);
 98             SEG::update(1, 1, b[0], x[i], y[i], 1);            
 99         }
100         int res = 0;
101         for (int i = 1; i <= b[0]; ++i)
102             ans[i] = SEG::query(1, 1, b[0], i);
103         for (int i = 1; i <= b[0]; ++i)
104         {
105             for (auto it : in[i])
106                 SEG::update(1, 1, b[0], x[it], y[it], -1);
107             res = max(res, ans[i] + SEG::a[1].Max);
108             for (auto it : out[i])
109                 SEG::update(1, 1, b[0], x[it], y[it], 1);
110         }
111         printf("%d\n", res);
112     }
113     return 0;
114 }

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