hdu3060Area2(任意多边形相交面积)

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多边形的面积求解是通过选取一个点（通常为原点或者多边形的第一个点）和其它边组成的三角形的有向面积。

对于两个多边形的相交面积就可以通过把多边形分解为三角形，求出三角形的有向面积递加。三角形为凸多边形，因此可以直接用凸多边形相交求面积的模板。

凸多边形相交后的部分肯定还是凸多边形，所以只需要判断哪些点是相交部分上的点，最后求下面积。

  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 510

 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) {} //构造函数 方便代码编写

 21 }p[N],q[N];

 22 typedef point pointt;

 23 pointt operator + (point a,point b)

 24 {

 25     return point(a.x+b.x,a.y+b.y);

 26 }

 27 pointt operator - (point a,point b)

 28 {

 29     return point(a.x-b.x,a.y-b.y);

 30 }

 31 int dcmp(double x)

 32 {

 33     if(fabs(x)<eps) return 0;

 34     else return x<0?-1:1;

 35 }

 36 bool operator == (const point &a,const point &b)

 37 {

 38     return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;

 39 }

 40 double cross(point a,point b)

 41 {

 42     return a.x*b.y-a.y*b.x;

 43 }

 44 double Polyarea(point p[],int n)

 45 {

 46     if(n<3) return 0;

 47     double area = 0;

 48     for(int i = 0 ; i < n ; i++)

 49     area+=cross(p[i],p[i+1]);

 50     return fabs(area)/2;

 51 }

 52 //double Polyarea(point p[], int n)

 53 //{

 54 //    if(n < 3) return 0.0;

 55 //    double s = p[0].y * (p[n - 1].x - p[1].x);

 56 //    p[n] = p[0];

 57 //    for(int i = 1; i < n; ++ i)

 58 //        s += p[i].y * (p[i - 1].x - p[i + 1].x);

 59 //    return fabs(s * 0.5);

 60 //}

 61 bool intersection1(point p1, point p2, point p3, point p4, point& p)      // 直线相交

 62 {

 63     double a1, b1, c1, a2, b2, c2, d;

 64     a1 = p1.y - p2.y;

 65     b1 = p2.x - p1.x;

 66     c1 = p1.x*p2.y - p2.x*p1.y;

 67     a2 = p3.y - p4.y;

 68     b2 = p4.x - p3.x;

 69     c2 = p3.x*p4.y - p4.x*p3.y;

 70     d = a1*b2 - a2*b1;

 71     if (!dcmp(d))    return false;

 72     p.x = (-c1*b2 + c2*b1) / d;

 73     p.y = (-a1*c2 + a2*c1) / d;

 74     return true;

 75 }

 76 

 77 double convexpolygon(point p[],point q[],int n,int m)

 78 {

 79     int i,j;

 80     point ch[20],cnt[20];

 81     p[n] = p[0],q[m] = q[0];

 82     memcpy(ch,q,sizeof(point)*(m+1));

 83     int f2,g = 0 ;

 84     for(i = 0 ;i < n; i++)

 85     {

 86         int f1 = dcmp(cross(p[i+1]-p[i],ch[0]-p[i]));

 87         g = 0 ;

 88         for(j = 0 ;j < m; j++,f1 = f2)

 89         {

 90             if(f1>=0) cnt[g++] = ch[j];

 91             f2 = dcmp(cross(p[i+1]-p[i],ch[j+1]-p[i]));

 92             if((f1^f2)==-2)

 93             {

 94                 point pp;

 95                 intersection1(p[i],p[i+1],ch[j],ch[j+1],pp);

 96                 cnt[g++] = pp;

 97             }

 98         }

 99         cnt[g] = cnt[0];

100         memcpy(ch,cnt,sizeof(point)*(g+1));

101         m = g;

102     }

103     return Polyarea(ch,g);

104 }

105 double solve(point p[],point q[],int n,int m)

106 {

107     int i,j;

108     double area = 0;

109     point tp[10],tq[10];

110     tp[0] = p[0];tq[0] = q[0];

111     for(i = 1 ;i < n-1; i++)

112     {

113         int k1 = dcmp(cross(p[i]-p[0],p[i+1]-p[0]));

114         tp[1] = p[i],tp[2] = p[i+1];

115         if(k1 < 0) swap(tp[1],tp[2]);

116 

117         for(j = 1 ;j < m-1 ;j++)

118         {

119             tq[1] = q[j],tq[2] = q[j+1];

120             int k2 = dcmp(cross(q[j]-q[0],q[j+1]-q[0]));

121             if(k2<0) swap(tq[1],tq[2]);

122 

123             area+=convexpolygon(tp,tq,3,3)*k1*k2;

124         }

125     }

126     return Polyarea(p,n)+Polyarea(q,m)-area;

127 }

128 int main()

129 {

130     int n,m,i;

131     while(scanf("%d%d",&n,&m)!=EOF)

132     {

133         for(i = 0; i < n ;i++)

134         {

135             scanf("%lf%lf",&p[i].x,&p[i].y);

136         }

137         for(i = 0; i < m; i++)

138         {

139             scanf("%lf%lf",&q[i].x,&q[i].y);

140         }

141         p[n] = p[0],q[m] = q[0];

142         double ans = solve(p,q,n,m);

143         printf("%.2f\n",ans);

144     }

145     return 0;

146 }

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