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

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。

 /*
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 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 typedef struct Point {
     int x, y;
     Point() {}
     Point(int xx, int yy) : x(xx), y(yy) {}
     bool operator==(Point p) {
         return x == p.x && y == p.y;
     }
     bool operator<(Point p) {
         if(x == p.x) return y < p.y;
         return x < p.x;
     }
 }Point;
 typedef struct Line {
     Point a, b;
     Line() {}
     Line(Point aa, Point bb) : a(aa), b(bb) {}
 }Line;
 const int maxn = ;
 const int maxm = ;
 Line line[maxn];
 Point s, e;
 int n, m;
 int tmp[maxm];
 int ans[maxm];

 int ok(Point p, Line l) {
     RT ((l.b.y-l.a.y)*(p.x-l.a.x)-(p.y-l.a.y)*(l.b.x-l.a.x));
 }

 bool cmp(Line a, Line b) {
     if(a.a == b.a) return a.b < b.b;
     return a.a < b.a;
 }

 int main() {
     // FRead();
     int x1, x2;
     bool flag = ;
     while(~Rint(n) && n) {
         Cls(tmp); Cls(ans);
         Rint(m); Rint(s.x); Rint(s.y); Rint(e.x); Rint(e.y);
         Rep(i, n) {
             Rint(x1); Rint(x2);
             line[i] = Line(Point(x1, s.y), Point(x2, e.y));
         }
         line[n] = Line(Point(e.x, s.y), e);
         Point p;
         sort(line, line+n+, cmp);
         W(m) {
             Rint(p.x); Rint(p.y);
             int l = , r = n;
             int ret;
             while(l <= r) {
                 int m = (l + r) >> ;
                 if(ok(p, line[m]) > ) {
                     ret = m;
                     r = m - ;
                 }
                 else l = m + ;
             }
             tmp[ret]++;
         }
         int hi = ;
         Rep(i, n+) {
             hi = max(hi, tmp[i]);
             ans[tmp[i]]++;
         }
         printf("Box\n");
         For(i, , hi+) {
             if(ans[i]) printf("%d: %d\n", i, ans[i]);
         }
     }
     RT ;
 }