链接：http://www.lydsy.com/JudgeOnline/problem.php?id=2618
题意：求出几个封闭图形围成的内部区域面积。
把每一条边作为有向直线，逆时针遍历全图，左侧的半平面交
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 int n,cnt; 8 const int maxn=55; 9 struct point 10 { 11 double x,y; 12 }p[maxn],a[maxn*maxn]; 13 double operator *(const point &a,const point &b) 14 { 15 return a.x*b.y-a.y*b.x; 16 } 17 point operator -(const point &a,const point &b) 18 { 19 return (point){a.x-b.x,a.y-b.y}; 20 } 21 struct line 22 { 23 point a,b;double slop; 24 friend bool operator <(const line &a,const line &b) 25 { 26 if(a.slop!=b.slop)return a.slop<b.slop; 27 return (b.b-a.a)*(a.b-a.a)<0; 28 } 29 void print() 30 { 31 printf("%lf %lf %lf %lf ",a.x,a.y,b.x,b.y); 32 } 33 }l[maxn*maxn],q[maxn*maxn]; 34 void pre() 35 { 36 for(int i=1;i<=cnt;i++)l[i].slop=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x); 37 sort(l+1,l+cnt+1);int tot=0; 38 for(int i=1;i<=cnt;i++) 39 { 40 if(l[i].slop!=l[i-1].slop)tot++; 41 l[tot]=l[i]; 42 } 43 cnt=tot; 44 } 45 point inter(line a,line b) 46 { 47 double k1,k2; 48 k1=(a.a-b.a)*(a.a-b.b),k2=(a.b-b.b)*(a.b-b.a); 49 k1=k1/(k1+k2); 50 point ans; 51 ans.x=a.a.x+k1*(a.b.x-a.a.x),ans.y=a.a.y+k1*(a.b.y-a.a.y); 52 return ans; 53 } 54 bool judge(point a,line b) 55 { 56 return (b.a-a)*(b.b-a)<0; 57 } 58 void work() 59 { 60 int L=1,R=0; 61 q[++R]=l[1],q[++R]=l[2]; 62 for(int i=3;i<=cnt;i++) 63 { 64 while(L<R&&judge(inter(q[R],q[R-1]),l[i]))R--; 65 while(L<R&&judge(inter(q[L],q[L+1]),l[i]))L++; 66 q[++R]=l[i]; 67 } 68 while(L<R&&judge(inter(q[R],q[R-1]),q[L]))R--; 69 cnt=0; 70 for(int i=L;i<R;i++)a[++cnt]=inter(q[i],q[i+1]); 71 a[++cnt]=inter(q[R],q[L]); 72 } 73 void getans() 74 { 75 double ans=0; 76 for(int i=1;i<cnt;i++)ans+=a[i]*a[i+1]; 77 ans+=a[cnt]*a[1]; 78 ans=fabs(ans/2.0); 79 printf("%0.3lf",ans); 80 } 81 int haha() 82 { 83 scanf("%d",&n); 84 for(int i=1;i<=n;i++) 85 { 86 int m;scanf("%d",&m); 87 for(int j=1;j<=m;j++) 88 { 89 scanf("%lf%lf",&p[j].x,&p[j].y); 90 if(j==1)continue; 91 l[++cnt].a=p[j-1];l[cnt].b=p[j]; 92 } 93 l[++cnt].a=p[m];l[cnt].b=p[1]; 94 } 95 pre();work();getans(); 96 } 97 int sb=haha(); 98 int main(){;}
就是所求的结果。