[POJ2398]Toy Storage（计算几何，二分，判断点在线段的哪一侧）

Posted 2020-07-17

题目链接：http://poj.org/problem?id=2398

思路RT，和POJ2318一样，就是需要排序，输出也不一样。手工画一下就明白了。注意叉乘的时候a×b是判断a在b的顺时针还是逆时针侧，>0是顺时针测，<0是逆时针侧，本题对应看成右、左侧，特别注意。

  1 /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%lld", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef map<string, int> msi;
 65 typedef vector<int> vi;
 66 typedef vector<LL> vl;
 67 typedef vector<vl> vvl;
 68 typedef vector<bool> vb;
 69 
 70 typedef struct Point {
 71     int x, y;
 72     Point() {}
 73     Point(int xx, int yy) : x(xx), y(yy) {}
 74     bool operator==(Point p) {
 75         return x == p.x && y == p.y;
 76     }
 77     bool operator<(Point p) {
 78         if(x == p.x) return y < p.y;
 79         return x < p.x;
 80     }
 81 }Point;
 82 typedef struct Line {
 83     Point a, b;
 84     Line() {}
 85     Line(Point aa, Point bb) : a(aa), b(bb) {}
 86 }Line;
 87 const int maxn = 5050;
 88 const int maxm = 5050;
 89 Line line[maxn];
 90 Point s, e;
 91 int n, m;
 92 int tmp[maxm];
 93 int ans[maxm];
 94 
 95 int ok(Point p, Line l) {
 96     RT ((l.b.y-l.a.y)*(p.x-l.a.x)-(p.y-l.a.y)*(l.b.x-l.a.x));
 97 }
 98 
 99 bool cmp(Line a, Line b) {
100     if(a.a == b.a) return a.b < b.b;
101     return a.a < b.a;
102 }
103 
104 int main() {
105     // FRead();
106     int x1, x2;
107     bool flag = 1;
108     while(~Rint(n) && n) {
109         Cls(tmp); Cls(ans);
110         Rint(m); Rint(s.x); Rint(s.y); Rint(e.x); Rint(e.y);
111         Rep(i, n) {
112             Rint(x1); Rint(x2);
113             line[i] = Line(Point(x1, s.y), Point(x2, e.y));
114         }
115         line[n] = Line(Point(e.x, s.y), e);
116         Point p;
117         sort(line, line+n+1, cmp);
118         W(m) {
119             Rint(p.x); Rint(p.y);
120             int l = 0, r = n;
121             int ret;
122             while(l <= r) {
123                 int m = (l + r) >> 1;
124                 if(ok(p, line[m]) > 0) {
125                     ret = m;
126                     r = m - 1;
127                 }
128                 else l = m + 1;
129             }
130             tmp[ret]++;
131         }
132         int hi = 0;
133         Rep(i, n+1) {
134             hi = max(hi, tmp[i]);
135             ans[tmp[i]]++;
136         }
137         printf("Box\n");
138         For(i, 1, hi+1) {
139             if(ans[i]) printf("%d: %d\n", i, ans[i]);
140         }
141     }
142     RT 0;
143 }