practice
Consider the left side represents its half-plane with radiation (a point and a vector)
Then we can sort the x-axis positive press angle axle (available atan2 (y, x) to achieve), and then deque rays to maintain the current in the cross
The reason to use double-ended queue, because the newly inserted a half-plane when the head of the queue and the tail are likely to be ejected, but also to be noted that the first bomb bomb squad first and then the tail
At the end of the first team and then also some of the tail of the bomb
example
luogu4196 convex polygon
1 #include<bits/stdc++.h> 2 #define pa pair<int,int> 3 #define CLR(a,x) memset(a,x,sizeof(a)) 4 #define MP make_pair 5 #define fi first 6 #define se second 7 using namespace std; 8 typedef long long ll; 9 typedef unsigned long long ull; 10 typedef unsigned int ui; 11 typedef long double ld; 12 const int maxl=2333,maxn=12,maxm=55; 13 const ld eps=1e-9; 14 15 inline char gc(){ 16 return getchar(); 17 static const int maxs=1<<16;static char buf[maxs],*p1=buf,*p2=buf; 18 return p1==p2&&(p2=(p1=buf)+fread(buf,1,maxs,stdin),p1==p2)?EOF:*p1++; 19 } 20 inline ll rd(){ 21 ll x=0;char c=gc();bool neg=0; 22 while(c<'0'||c>'9'){if(c=='-') neg=1;c=gc();} 23 while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+c-'0',c=gc(); 24 return neg?(~x+1):x; 25 } 26 27 struct Node{ 28 ld x,y; 29 Node(ld _x=0,ld _y=0){x=_x,y=_y;} 30 }p[maxn][maxm],it[maxl]; 31 struct Line{ 32 Node x,y; 33 Line(Node _x=0,Node _y=0){x=_x,y=_y;} 34 }l[maxl]; 35 int N,M[maxn],cnt; 36 int hd,tl,q[maxl]; 37 38 inline ld operator ^(const Node a,const Node b){return a.x*b.y-a.y*b.x;} 39 inline Node operator +(const Node a,const Node b){return Node(a.x+b.x,a.y+b.y);} 40 inline Node operator -(const Node a,const Node b){return Node(a.x-b.x,a.y-b.y);} 41 inline Node operator *(const Node a,const ld b){return Node(a.x*b,a.y*b);} 42 inline ld myabs(const ld x){return x>0?x:-x;} 43 inline Node inter(const Line a,const Line b){ 44 ld k1=(a.y-a.x)^(b.y-a.x),k2=(b.x-a.x)^(a.y-a.x); 45 return b.x+(b.y-b.x)*(k2/(k1+k2)); 46 } 47 inline bool onright(const Line a,const Node b){return ((b-a.x)^(a.y-a.x))>=-eps;} 48 inline ld slope(Line a){return atan2((a.y-a.x).y,(a.y-a.x).x);} 49 50 inline bool cmp(Line a,Line b){ 51 ld x=slope(a),y=slope(b); 52 return (myabs(x-y)<eps)?(!onright(b,a.x)):(x<y); 53 } 54 55 inline ld solve(){ 56 sort(l+1,l+cnt+1,cmp); 57 hd=1,tl=0; 58 for(int i=1;i<=cnt;i++){ 59 // printf("~%Lf %Lf\n",(l[i].y-l[i].x).x,(l[i].y-l[i].x).y); 60 if(i>1&&myabs(slope(l[i])-slope(l[i-1]))<eps) continue; 61 while(hd<tl&&onright(l[i],inter(l[q[tl]],l[q[tl-1]]))) tl--; 62 while(hd<tl&&onright(l[i],inter(l[q[hd]],l[q[hd+1]]))) hd++; 63 q[++tl]=i; 64 } 65 while(hd<tl&&onright(l[q[hd]],inter(l[q[tl]],l[q[tl-1]]))) tl--; 66 // while(hd<tl&&onright(l[q[tl]],inter(l[q[hd]],l[q[hd+1]]))) hd++; 67 // for(int i=hd;i<=tl;i++) printf("~%Lf %Lf %Lf %Lf\n",l[q[i]].x.x,l[q[i]].x.y,l[q[i]].y.x,l[q[i]].y.y); 68 if(hd+2>tl) return 0; 69 ld re=0; 70 q[tl+1]=q[hd];int n=0; 71 for(int i=hd;i<=tl;i++) it[++n]=inter(l[q[i]],l[q[i+1]]); 72 for(int i=2;i<n;i++) re+=myabs((it[i]-it[1])^(it[i+1]-it[1])); 73 return re/2; 74 } 75 76 int main(){ 77 //freopen("","r",stdin); 78 N=rd(); 79 for(int i=1;i<=N;i++){ 80 M[i]=rd(); 81 for(int j=0;j<M[i];j++) 82 p[i][j].x=rd(),p[i][j].y=rd(); 83 for(int j=0;j<M[i];j++){ 84 l[++cnt]=Line(p[i][j],p[i][(j+1)%M[i]]); 85 } 86 } 87 printf("%.3Lf\n",solve()); 88 return 0; 89 }