BZOJ2618 [Cqoi2006]凸多边形

那个叫啥，半平面交。。。

第一次写于是只能按照惯例，orz hzwer去~~~

把一个凸多边形搞成好多条线段，于是题目就变成了一堆线段的半平面交。。。

怎么感觉比仙人掌还简单一点的说。。。就是有点长

  1 /**************************************************************
  2     Problem: 2618
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:0 ms
  7     Memory:932 kb
  8 ****************************************************************/
  9
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13
 14 using namespace std;
 15 typedef double lf;
 16 const int N = 1005;
 17 struct Point{
 18     lf x, y;
 19     Point(void){}
 20     Point(lf a, lf b) : x(a), y(b){}
 21 }p[N], a[N];
 22
 23 struct Line{
 24     Point a, b;
 25     lf slope;
 26     Line(void){}
 27     Line(Point x, Point y, lf z) : a(x), b(y), slope(z){}
 28 }l[N], q[N];
 29 int n, cnt, tot, num;
 30 lf ans;
 31
 32 inline int read(){
 33     int x = 0, sgn = 1;
 34     char ch = getchar();
 35     while (ch < ‘0‘ || ch > ‘9‘){
 36         if (ch == ‘-‘) sgn = -1;
 37         ch = getchar();
 38     }
 39     while (ch >= ‘0‘ && ch <= ‘9‘){
 40         x = x * 10 + ch - ‘0‘;
 41         ch = getchar();
 42     }
 43     return sgn * x;
 44 }
 45
 46 inline lf operator * (const Point a, const Point b){
 47     return a.x * b.y - a.y * b.x;
 48 }
 49
 50 inline Point operator - (const Point a, const Point b){
 51     return Point(a.x - b.x, a.y - b.y);
 52 }
 53
 54 inline bool operator < (const Line a, const Line b){
 55     return a.slope == b.slope ? (a.b - a.a) * (b.b - a.a) > 0 : a.slope < b.slope;
 56 }
 57
 58 inline Point inter(const Line a, const Line b){
 59     lf k1, k2, tmp;
 60     Point res;
 61     k1 = (b.b - a.a) * (a.b - a.a);
 62     k2 = (a.b - a.a) * (b.a - a.a);
 63     tmp = k1 / (k1 + k2);
 64     res = Point(b.b.x + tmp *(b.a.x - b.b.x), b.b.y + tmp * (b.a.y - b.b.y));
 65     return res;
 66 }
 67
 68 inline bool check(const Line a, const Line b, const Line t){
 69     Point p = inter(a, b);
 70     return (t.b - t.a) * (p - t.a) < 0;
 71 }
 72
 73 void work(){
 74     sort(l + 1, l + cnt + 1);
 75     int L = 1, R = 2;
 76     tot = 0;
 77     for (int i = 1; i <= cnt; ++i){
 78         if (l[i].slope != l[i -1].slope) ++tot;
 79         l[tot] = l[i];
 80     }
 81
 82     cnt = tot, tot = 0;
 83     q[1] = l[1], q[2] = l[2];
 84     for (int i = 3; i <= cnt; ++i){
 85         while (L < R && check(q[R - 1], q[R], l[i])) --R;
 86         while (L < R && check(q[L + 1], q[L], l[i])) ++L;
 87         q[++R] = l[i];
 88     }
 89     while (L < R && check(q[R - 1], q[R], q[L])) --R;
 90     while (L < R && check(q[L + 1], q[L], q[R])) ++L;
 91     q[R + 1] = q[L];
 92     for (int i = L; i <= R; ++i)
 93         a[++tot] = inter(q[i], q[i + 1]);
 94 }
 95
 96 void get_ans(){
 97     ans = 0;
 98     if (tot < 3) return;
 99     a[++tot] = a[1];
100     for (int i = 1; i <= tot; ++i)
101         ans += a[i] * a[i + 1];
102     ans = fabs(ans) / 2;
103 }
104
105 int main(){
106     n = read();
107     int X, Y;
108     for (int i = 1; i <= n; ++i){
109         num = read();
110         for (int j = 1; j <= num; ++j){
111             X = read(), Y = read();
112             p[j] = Point(X, Y);
113         }
114         p[num + 1] = p[1];
115         for (int j = 1; j <= num; ++j)
116             l[++cnt] = Line(p[j], p[j + 1], 0);
117     }
118     for (int i = 1; i <= cnt; ++i)
119         l[i].slope = atan2(l[i].b.y - l[i].a.y, l[i].b.x - l[i].a.x);
120     work();
121     get_ans();
122     printf("%.3lf\n", ans);
123     return 0;
124 }