【模板】【poj1279】 计几半平面交求多边形内核 nlogn
题目链接：https://vjudge.net/problem/POJ-1279
参考：https://blog.csdn.net/commonc/article/details/55260747
以及：https://blog.csdn.net/commonc/article/details/55252420
题意：求多边形内核的面积。
都在注释里了。nlogn半平面交模板
1 /************************************************************************* 2 > File Name: poj1279.cpp 3 # File Name: poj1279.cpp 4 # Author : xiaobuxie 5 # QQ : 760427180 6 # Email:760427180@qq.com 7 # Created Time: 2019年09月26日 星期四 16时12分12秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<cstdio> 12 #include<map> 13 #include<cmath> 14 #include<cstring> 15 #include<set> 16 #include<queue> 17 #include<vector> 18 #include<algorithm> 19 using namespace std; 20 typedef long long ll; 21 #define inf 0x3f3f3f3f 22 #define pq priority_queue<int,vector<int>,greater<int> > 23 ll gcd(ll a,ll b){ 24 if(a<b) return gcd(b,a); 25 return b==0?a:gcd(b,a%b); 26 } 27 const int N = 1500+9; 28 #define eps 1e-8 29 30 31 32 //q和L代表原来点和原来向量 33 //quel代表队列里的向量,queq[i]代表quel[i] 和 quel[i+1]的交点 34 int n,h,t; 35 struct Point{ 36 double x,y; 37 Point operator - (const Point & b)const{ 38 return (Point){x-b.x,y-b.y}; 39 } 40 Point operator + (const Point& b)const{ 41 return (Point){x+b.x , y+b.y}; 42 } 43 Point operator * (double b)const{ 44 return (Point){x*b,y*b}; 45 } 46 double operator ^ (const Point& b)const{ 47 return x*b.y-b.x*y; 48 } 49 }p[N],quep[N]; 50 // Line是有向向量，ang代表和x轴的角度，选的半平面在向量的右面，即交点在向量左边的舍去 51 struct Line{ 52 Point s,t; 53 double ang; 54 //注意ang相同时 55 bool operator < (const Line& b)const{ 56 if(ang != b.ang) return ang < b.ang; 57 return ( (b.t - s) ^ (b.s - s) ) >0; 58 } 59 }L[N],quel[N]; 60 61 int sgn(double x){ 62 if(fabs(x) < eps) return 0; 63 if(x > 0) return 1; 64 return -1; 65 } 66 // clock wise判断是否为顺时针，要把点弄成顺时针(这里处理点是顺或逆的),如果乱序，直接极角排序 67 bool cw(){ 68 double res=0; 69 for(int i=2; i<n ; ++i){ 70 res += ( (p[i]-p[1]) ^ (p[i+1]-p[1]) ); 71 } 72 return sgn(res) < 0; 73 } 74 //判断点是否在向量左边 75 bool onleft(Point u, Line a){ 76 return sgn( (a.s-u) ^ (a.t-u) ) > 0; 77 } 78 //两直线交点 79 Point line_inter(Line l1,Line l2){ 80 double b = ((l1.t-l1.s)^(l2.s-l1.s))/((l2.t-l2.s)^(l1.t-l1.s)); 81 return l2.s + (l2.t - l2.s)*b; 82 } 83 void init(){ 84 scanf("%d",&n); 85 for(int i=1;i<=n;++i) scanf("%lf %lf",&p[i].x,&p[i].y); 86 if( !cw() ) reverse(p+1,p+1+n); 87 p[n+1]=p[1]; 88 for(int i=1;i<=n;++i) L[i]=(Line){ p[i], p[i+1], atan2(p[i+1].y-p[i].y,p[i+1].x-p[i].x)}; 89 sort(L+1,L+1+n); 90 int nn=1; 91 for(int i=2;i<=n;++i){ 92 if(L[i].ang != L[i-1].ang) L[++nn]=L[i]; 93 } 94 n = nn; 95 } 96 //处理半平面交 97 void solve(){ 98 h = t = 1; 99 quel[1] = L[1]; 100 for(int i = 2 ; i<=n ; ++i ){ 101 while( h<t && onleft( quep[t-1] , L[i] ) ) --t; 102 while( h<t && onleft( quep[h], L[i]) ) ++h; 103 quel[++t] = L[i]; 104 if( h<t ) quep[t-1] = line_inter(quel[t-1],quel[t]); 105 } 106 while( h<t && onleft( quep[t-1] , quel[h] ) ) --t; 107 quep[t] = line_inter(quel[h],quel[t]); 108 } 109 //求多边形内核的面积 110 void solve_area(){ 111 double area=0; 112 for(int i=h+1; i<t ; ++i){ 113 area += ( (quep[i] - quep[h]) ^ (quep[i+1] - quep[h]) ); 114 } 115 area/=2; 116 printf("%.2f\n",area); 117 } 118 int main(){ 119 int T; scanf("%d",&T); 120 while(T--){ 121 init(); 122 solve(); 123 solve_area(); 124 } 125 return 0; 126 }
