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ecnu1624求交集多边形面积

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本来在刷hdu的一道题。。一直没过,看到谈论区发现有凹的,我这种方法只能过凸多边形的相交面积。。

就找来这道题试下水。

两个凸多边形相交的部分要么没有 要么也是凸多边形,那就可以把这部分单独拿出来极角排序、叉积求面积。这部分的顶点要么p在q内的顶点,要么是q在p内的顶点,要么是两凸多边形的交点。

用到了点在多边形内的判定模板。

  1 #include <iostream>  2 #include<cstdio>  3 #include<cstring>  4 #include<algorithm>  5 #include<stdlib.h>  6 #include<vector>  7 #include<cmath>  8 #include<queue>  9 #include<set> 10 using namespace std; 11 #define N 10000 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct point 18 { 19     double x,y; 20     point(double x=0,double y =0 ):x(x),y(y) {} 21 } p[N],q[N],ch[N],chh[N]; 22 typedef point pointt; 23 point operator -(point a,point b) 24 { 25     return point(a.x-b.x,a.y-b.y); 26 } 27 int dcmp(double x) 28 { 29     if(fabs(x)<eps) return 0; 30     return x<0?-1:1; 31 } 32 double cross(point a,point b) 33 { 34     return a.x*b.y-a.y*b.x; 35 } 36 double dis(point a) 37 { 38     return sqrt(a.x*a.x+a.y*a.y); 39 } 40 double getarea(point p[],int n) 41 { 42     int i; 43     double area = 0; 44     for(i =0  ; i < n-1; i++) 45         area+=cross(p[i]-p[0],p[i+1]-p[0]); 46     area = fabs(area)/2; 47     return area; 48 } 49 bool PtInPolygon (point p, point ptPolygon[], int nCount) 50 { 51     int i,nCross = 0; 52     for(i =0 ; i< nCount ; i++) 53     { 54         point p1 = ptPolygon[i]; 55         point p2 = ptPolygon[(i+1)%nCount]; 56         if(dcmp(p1.y-p2.y)==0) continue; 57         if(dcmp(p.y-min(p1.y,p2.y))<0) continue; 58         if(dcmp(p.y-max(p1.y,p2.y))>=0) continue; 59         double x = (double)(p.y-p1.y)*(double)(p2.x-p1.x)/(double)(p2.y-p1.y)+p1.x; 60         if(x>p.x)   nCross++; 61     } 62     return (nCross % 2 == 1); 63 } 64 bool segprointer(point a1,point a2,point b1,point b2) 65 { 66     double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1); 67     double c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1); 68     return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0; 69 } 70 bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交 71 { 72     double a1, b1, c1, a2, b2, c2, d; 73     a1 = p1.y - p2.y; 74     b1 = p2.x - p1.x; 75     c1 = p1.x*p2.y - p2.x*p1.y; 76     a2 = p3.y - p4.y; 77     b2 = p4.x - p3.x; 78     c2 = p3.x*p4.y - p4.x*p3.y; 79     d = a1*b2 - a2*b1; 80     if (!dcmp(d))    return false; 81     p.x = (-c1*b2 + c2*b1) / d; 82     p.y = (-a1*c2 + a2*c1) / d; 83     return true; 84 } 85 double mul(point p0,point p1,point p2) 86 { 87     return cross(p1-p0,p2-p0); 88 } 89   90 bool cmp(point a,point b) 91 { 92     if(dcmp(mul(ch[0],a,b))==0) 93         return dis(a-ch[0])<dis(b-p[0]); 94     else 95         return dcmp(mul(ch[0],a,b))>0; 96 } 97 bool cmpp(point a,point b) 98 { 99     if(dcmp(a.x-b.x)==0) return a.y<b.y;100     return a.x<b.x;101 }102 int main()103 {104     int n,m,i,j;105     while(scanf("%d",&n)!=EOF)106     {107  108         for(i = 0; i < n ; i++)109             scanf("%lf%lf",&p[i].x,&p[i].y);110         p[n] = p[0];111         scanf("%d",&m);112         for(i = 0; i < m; i++)113             scanf("%lf%lf",&q[i].x,&q[i].y);114         q[m] = q[0];115         int g = 0;116         for(i = 0; i < n ; i ++)117         {118             if(PtInPolygon(p[i],q,m))119                 ch[g++] = p[i];120         }121         for(i = 0; i < m ; i ++)122         {123             if(PtInPolygon(q[i],p,n))124                 ch[g++] = q[i];125         }126         for(i = 0 ; i < n;  i++)127         {128             for(j = 0; j < m ; j++)129             {130                 if(segprointer(p[i],p[i+1],q[j],q[j+1])==0) continue;131                 point pp;132                 if(!intersection1(p[i],p[i+1],q[j],q[j+1],pp))continue;133                 ch[g++] = pp;134             }135         }136         double ans = 0;//getarea(p,n)+getarea(q,m);137         if(g==0)138             ans = 0;139         else140         {141             int k = 0,o=0;142             sort(ch,ch+g,cmpp);143             chh[o++] = ch[0];144             for(i  = 1; i < g; i++)145             if(dcmp(ch[i].x-ch[i-1].x)==0&&dcmp(ch[i].y-ch[i-1].y)==0)146             continue;147             else chh[o++] = ch[i];148             g = o;149             for(i = 1; i < g ; i ++)150             {151                 if(dcmp(chh[i].y-chh[k].y)<0||(dcmp(chh[i].y-chh[k].y)==0&&dcmp(chh[i].x-chh[k].x)<0))152                     k = i;153             }154             if(k!=0) swap(chh[k],chh[0]);155             sort(chh+1,chh+g,cmp);156             ans = getarea(chh,g);157         }158         printf("%.2f\n",ans);159     }160     return 0;161 }
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