我们先看每个点可能从哪些点折过来的，2^10枚举对角线是否用到。
然后再模拟折法，查看每个点是否满足要求。
恩，计算几何比较恶心，还好前几天刚写过一道更恶心的计算几何，点类直接拷过来2333。
1 /************************************************************** 2 Problem: 1074 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:24 ms 7 Memory:980 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <algorithm> 13 14 using namespace std; 15 typedef double lf; 16 17 const int N = 10; 18 const lf eps = 1e-8; 19 20 int n, cnt; 21 22 inline lf sqr(lf x) { 23 return x * x; 24 } 25 26 inline int dcmp(lf x) { 27 return fabs(x) <= eps ? 0 : (x > eps ? 1 : -1); 28 } 29 30 struct point { 31 lf x, y; 32 point() {} 33 point(lf _x, lf _y) : x(_x), y(_y) {} 34 35 inline point operator + (point p) { 36 return point(x + p.x, y + p.y); 37 } 38 inline point operator - (point p) { 39 return point(x - p.x, y - p.y); 40 } 41 inline lf operator * (point p) { 42 return x * p.y - y * p.x; 43 } 44 inline lf operator % (point p) { 45 return x * p.x + y * p.y; 46 } 47 inline point operator * (lf a) { 48 return point(x * a, y * a); 49 } 50 inline point operator / (lf a) { 51 return point(x / a, y / a); 52 } 53 54 inline bool operator < (const point &p) const { 55 return dcmp(x - p.x) == 0 ? dcmp(y - p.y) < 0 : dcmp(x - p.x) < 0; 56 } 57 inline bool operator != (const point &p) const { 58 return dcmp(x - p.x) || dcmp(y - p.y); 59 } 60 inline bool operator == (const point &p) const { 61 return !dcmp(x - p.x) && !dcmp(y - p.y); 62 } 63 64 inline void read_in() { 65 scanf("%lf%lf", &x, &y); 66 } 67 friend inline lf dis2(point p) { 68 return sqr(p.x) + sqr(p.y); 69 } 70 friend inline lf dis(point p) { 71 return sqrt(dis2(p)); 72 } 73 friend inline lf angle(point p, point q) { 74 return acos(p % q / dis(p) / dis(q)); 75 } 76 friend inline point rotate(point p, lf A) { 77 lf s = sin(A), c = cos(A); 78 return point(p.x * c - p.y * s, p.x * s + p.y * c); 79 } 80 } ans[N << 10]; 81 82 struct line { 83 point p, v; 84 line() {} 85 line(point _p, point _v) : p(_p), v(_v){} 86 } l[N]; 87 88 inline point reverse(point p, line l, int f) { 89 return l.p + rotate(p - l.p, angle(p - l.p, l.v) * 2 * f); 90 } 91 92 inline bool on_left(point p, line l) { 93 return dcmp((p - l.p) * l.v) < 0; 94 } 95 96 inline bool on_right(point p, line l) { 97 return dcmp((p - l.p) * l.v) > 0; 98 } 99 100 void dfs(point p, int d) { 101 ans[++cnt] = p; 102 if (d == 0) return; 103 dfs(p, d - 1); 104 if (on_left(p, l[d])) dfs(reverse(p, l[d], -1), d - 1); 105 } 106 107 inline bool check(lf x) { 108 return dcmp(x) > 0 && dcmp(x - 100) < 0; 109 } 110 111 inline bool check(point p) { 112 return check(p.x) && check(p.y); 113 } 114 115 inline point find(point p) { 116 int i; 117 for (i = 1; i <= n; ++i) 118 if (on_right(p, l[i])) p = reverse(p, l[i], 1); 119 else if (!on_left(p, l[i])) return point(-1, -1); 120 return p; 121 } 122 123 int work(point p) { 124 int res = 0, i; 125 cnt = 0; 126 dfs(p, n); 127 sort(ans + 1, ans + cnt + 1); 128 cnt = unique(ans + 1, ans + cnt + 1) - ans - 1; 129 for (i = 1; i <= cnt; ++i) 130 if (check(ans[i]) && find(ans[i]) == p) ++res; 131 return res; 132 } 133 134 int main() { 135 int i, Q; 136 point x, y; 137 scanf("%d", &n); 138 for (i = 1; i <= n; ++i) { 139 x.read_in(), y.read_in(); 140 l[i] = line(x, y - x); 141 } 142 scanf("%d", &Q); 143 while (Q--) { 144 x.read_in(); 145 printf("%d ", work(x)); 146 } 147 return 0; 148 }