半平面：一条直线把二维平面分成两个平面。
半平面交：在二维几何平面上，给出若干个半平面，求它们的公共部分
半平面交的结果：1.凸多边形（后面会讲解到）2.无界，因为有可能若干半平面没有形成封闭3.直线，线段，点，空（属于特殊情况吧）
算法：1：根据上图可以知道，运用给出的多边形每相邻两点形成一条直线来切割原有多边形，如果多边形上的点i在有向直线的左边或者在直线上即保存起来，否则判断此点的前一个点i-1和后一个点i+1是否在此直线的左边或线上，在的话分别用点i和点i-1构成的直线与此时正在切割的直线相交求出交点，这个交点显然也要算在切割后剩下的多边形里，同理点i和点i+1。原多边形有n条边，每条边都要进行切割，所以时间复杂度为O(n^2)。
2：第二种就是训练指南上面详细讲解的运用双端队列的半平面交算法，时间复杂度为O(nlogn)。仔细阅读代码应该能理解。
代码实现：以poj3130为例，裸的模板。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 #define exp 1e-10
9 #define PI 3.141592654
10 using namespace std;
11 const int maxn=111;
12 struct Point
13 {
14 double x,y;
15 Point (double x=0,double y=0):x(x),y(y){}
16 }an[maxn],bn[maxn],cn[maxn];
17 ///an:记录最开始的多边形；bn:临时保存新切割的多边形；cn:保存新切割出的多边形
18 typedef Point Vector;
19 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
20 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
21 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
22 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); }
23 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; }
24 double cross(Vector A,Vector B)
25 {
26 return A.x*B.y-B.x*A.y;
27 }
28
29 double A,B,C;
30 int n,m;
31 void getline(Point a,Point b)///获取直线 Ax + By + C = 0
32 {
33 A=b.y-a.y;
34 B=a.x-b.x;
35 C=b.x*a.y-a.x*b.y;
36 }
37 ///getline()函数得到的直线和点a和点b所连直线的交点
38 Point intersect(Point a,Point b)
39 {
40 double u=fabs(A*a.x+B*a.y+C);
41 double v=fabs(A*b.x+B*b.y+C);
42 Point ans;
43 ans.x=(a.x*v+b.x*u)/(u+v);
44 ans.y=(a.y*v+b.y*u)/(u+v);
45 return ans;
46 }
47 void cut()///切割，原多边形的点为顺时针存储
48 {
49 int cnt=0;
50 for (int i=1 ;i<=m ;i++)
51 {
52 if (A*cn[i].x + B*cn[i].y + C>=0) bn[++cnt]=cn[i];
53 else
54 {
55 if (A*cn[i-1].x + B*cn[i-1].y + C > 0) bn[++cnt]=intersect(cn[i-1],cn[i]);
56 if (A*cn[i+1].x + B*cn[i+1].y + C > 0) bn[++cnt]=intersect(cn[i+1],cn[i]);
57 }
58 }
59 for (int i=1 ;i<=cnt ;i++) cn[i]=bn[i];
60 cn[0]=bn[cnt];
61 cn[cnt+1]=bn[1];
62 m=cnt;///新切割出的多边形的点数
63 }
64 void solve()
65 {
66 for (int i=1 ;i<=n ;i++) cn[i]=an[i];
67 an[n+1]=an[1];
68 cn[n+1]=an[1];
69 cn[0]=an[n];
70 m=n;
71 for (int i=1 ;i<=n ;i++)
72 {
73 getline(an[i],an[i+1]);
74 cut();
75 }
76 }
77 int main()
78 {
79 while (scanf("%d",&n)!=EOF && n)
80 {
81 for (int i=1 ;i<=n ;i++) scanf("%lf%lf",&an[i].x,&an[i].y);
82 reverse(an+1,an+n+1);
83 solve();
84 if (m) puts("1");
85 else puts("0");
86 }
87 return 0;
88 }
poj3335，给出两种方法
1.O(n^2)
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 #define exp 1e-10
9 #define PI 3.141592654
10 using namespace std;
11 const int maxn=111;
12 struct Point
13 {
14 double x,y;
15 Point (double x=0,double y=0):x(x),y(y){}
16 }an[maxn],bn[maxn],cn[maxn];
17 typedef Point Vector ;
18 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
19 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
20 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
21 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); }
22 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; }
23 double cross(Vector A,Vector B)
24 {
25 return A.x*B.y-B.x*A.y;
26 }
27
28 int n,m;
29 double A,B,C;
30 void getline(Point a,Point b)
31 {
32 A=b.y-a.y;
33 B=a.x-b.x;
34 C=b.x*a.y-a.x*b.y;
35 }
36 Point intersect(Point a,Point b)
37 {
38 double u=fabs(A*a.x+B*a.y+C);
39 double v=fabs(A*b.x+B*b.y+C);
40 Point ans;
41 ans.x=(a.x*v+b.x*u)/(u+v);
42 ans.y=(a.y*v+b.y*u)/(u+v);
43 return ans;
44 }
45 void cut()
46 {
47 int cnt=0;
48 for (int i=1 ;i<=m ;i++)
49 {
50 if (A*cn[i].x+B*cn[i].y+C>=0)
51 bn[++cnt]=cn[i];
52 else
53 {
54 if (A*cn[i-1].x+B*cn[i-1].y+C>0)
55 bn[++cnt]=intersect(cn[i-1],cn[i]);
56 if (A*cn[i+1].x+B*cn[i+1].y+C>0)
57 bn[++cnt]=intersect(cn[i+1],cn[i]);
58 }
59 }
60 for (int i=1 ;i<=cnt ;i++) cn[i]=bn[i];
61 cn[0]=bn[cnt];
62 cn[cnt+1]=bn[1];
63 m=cnt;
64 }
65 void solve()
66 {
67 for (int i=1 ;i<=n ;i++) cn[i]=an[i];
68 an[n+1]=an[1];
69 cn[n+1]=cn[1];
70 cn[0]=cn[n];
71 m=n;
72 for (int i=1 ;i<=n ;i++)
73 {
74 getline(an[i],an[i+1]);
75 cut();
76 }
77 }
78 int main()
79 {
80 int t;
81 scanf("%d",&t);
82 while (t--)
83 {
84 scanf("%d",&n);
85 for (int i=1 ;i<=n ;i++) scanf("%lf%lf",&an[i].x,&an[i].y);
86 solve();
87 if (!m) printf("NO\n");
88 else printf("YES\n");
89 }
90 return 0;
91 }
2.O(nlogn)
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 #define exp 1e-10
9 #define PI 3.141592654
10 using namespace std;
11 const int maxn=111;
12 struct Point
13 {
14 double x,y;
15 Point (double x=0,double y=0):x(x),y(y){}
16 }an[maxn];
17 typedef Point Vector;
18 struct Line
19 {
20 Point p;
21 Vector v;
22 double ang;
23 Line (){}
24 Line (Point p,Vector v):p(p),v(v){ang=atan2(v.y,v.x); }
25 //Line (Point p,Vector v):p(p),v(v) {ang=atan2(v.y,v.x); }
26 friend bool operator < (Line a,Line b)
27 {
28 return a.ang<b.ang;
29 }
30 }bn[maxn];
31 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
32 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
33 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
34 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); }
35 int dcmp(double x) {if (fabs(x)<exp) return 0;return x>0 ? 1 : -1; }
36 double cross(Vector A,Vector B)
37 {
38 return A.x*B.y-B.x*A.y;
39 }
40 bool OnLeft(Line L,Point p)
41 {
42 return cross(L.v,p-L.p)>=0;///点P在有向直线L的左边（>=0说明在线上也算）
43 }
44 Point GetIntersection(Line a,Line b)
45 {
46 Vector u=a.p-b.p;
47 double t=cross(b.v,u)/cross(a.v,b.v);
48 return a.p+a.v*t;
49 }
50 //Point GetIntersection(Line l1, Line l2) {
51 // Point p;
52 // double dot1,dot2;
53 // //dot1 = multi(l2.a, l1.b, l1.a);
54 // dot1=cross(l1.b-l2.a , l1.a-l2.a);
55 // //dot2 = multi(l1.b, l2.b, l1.a);
56 // dot2=cross(l2.b-l1.b , l1.a-l1.b);
57 // p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot2 + dot1);
58 // p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot2 + dot1);
59 // return p;
60 //}
61 int HalfplaneIntersection(Line *L,int n,Point *poly)
62 {
63 sort(L,L+n);
64
65 int first,last;
66 Point *p=new Point[n];
67 Line *q=new Line[n];
68 q[first=last=0]=L[0];
69 for (int i=1 ;i<n ;i++)
70 {
71 while (first<last && !OnLeft(L[i],p[last-1])) last--;
72 while (first<last && !OnLeft(L[i],p[first])) first++;
73 q[++last]=L[i];
74 if (fabs(cross(q[last].v , q[last-1].v))<exp)
75 {
76 last--;
77 if (OnLeft(q[last] , L[i].p)) q[last]=L[i];
78 }
79 if (first<last) p[last-1]=GetIntersection(q[last-1],q[last]);
80 }
81 while (first<last && !OnLeft(q[first],p[last-1])) last--;
82 if (last-first<=1) return 0;
83 p[last]=GetIntersection(q[last],q[first]);
84 int m=0;
85 for (int i=first ;i<=last ;i++) poly[m++]=p[i];
86 return m;
87 }
88 void calPolygon(Point *p, int n, double &area, bool &shun)
89 {
90 p[n] = p[0];
91 area = 0;
92 double tmp;
93 for (int i = 0; i < n; i++)
94 area += p[i].x * p[i + 1].y - p[i].y * p[i + 1].x;
95 area /= 2.0;
96 if (shun = area < 0)
97 area = -area;
98 }
99 bool calCore(Point *ps, int n)
100 {
101 Line l[maxn];
102 ps[n] = ps[0];
103 bool shun;
104 double area;
105 calPolygon(ps, n, area, shun);
106 if (shun)
107 for (int i = 0; i < n; i++)
108 bn[i] = Line(ps[i], ps[i] - ps[i + 1]);
109 else
110 for (int i = 0; i < n; i++)
111 bn[i] = Line(ps[i], ps[i + 1] - ps[i]);
112 Point pp[maxn];
113 return HalfplaneIntersection(bn, n, pp);
114 }
115 int main()
116 {
117 int t;
118 int n;
119 scanf("%d",&t);
120 while (t--)
121 {
122 scanf("%d",&n);
123 Point cn[maxn];
124 for (int i=0 ;i<n ;i++)
125 {
126 scanf("%lf%lf",&cn[i].x,&cn[i].y);
127 }
128 // reverse(cn,cn+n);
129 // for (int i=0 ;i<n ;i++)
130 // {
131 // bn[i].p=cn[i];
132 // bn[i].v=cn[(i+1)%n]-cn[i];
133 // bn[i].ang=atan2(bn[i].v.y , bn[i].v.x);
134 // }
135 // int m=HalfplaneIntersection(bn,n,an);
136 if (!calCore(cn,n)) puts("NO");
137 else puts("YES");
138 }
139 return 0;
140 }
练习：
后续：欢迎提出宝贵的意见。
时间： 2024-10-05 04:09:39