You n points, find the largest convex polygon, so that this is not the point convex polygon.
After considering the order graham algorithm for convex hull, it is necessary to find the lower left corner, and then sort the polar angle sweep point, first enumerate the lower left corner.
This process, if you encounter interior point, we need to put this in the point excluded, but now need to point out is excluded.
Since the convex hull boundary uncertain, dp processing needs, starting two sides a triangle as a boundary line, and then continue the enumeration of possible boundary extension, while ensuring not contain the point.
#include<iostream> #include<cstdio> #include<string> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include <set> #include <queue> #define ll long long #define ld long double #define lson l,m,rt<<1 #define pi acos(-1) #define rson m+1,r,rt<<1|1 #define fo(i,l,r) for(int i = l;i <= r;i++) #define fd(i,l,r) for(int i = r;i >= l;i--) #define mem(x) memset(x,0,sizeof(x)) #define eps 3e-11 using namespace std; const int maxn = 55; const ll inf = 1e9; const ll mod = 998244353; ll read() { ll x=0,f=1; char ch=getchar(); while(!(ch>='0'&&ch<='9')) { if(ch=='-')f=-1; ch=getchar(); }; while(ch>='0'&&ch<='9') { x=x*10+(ch-'0'); ch=getchar(); }; return x*f; } int sgn(double x){ if(fabs(x)<eps)return 0; if(x<0)return -1; return 1; } struct Point{ int id; double x,y; Point(){ } Point(double _x,double _y){ x=_x; y=_y; } friend bool operator < (Point a,Point b) { return sgn(a.x-b.x)==0?sgn(a.y-b.y)<0:a.x<b.x; } double operator ^ (const Point &b) const{ return x*b.y - y*b.x; } bool operator == (const Point &b) const{ return sgn(x-b.x)==0&&sgn(y-b.y)==0; } Point operator - (const Point &b) const{ return Point(x-b.x,y-b.y); } double distance(Point b){ return sqrt((x-b.x)*(x-b.x) + (y-b.y)*(y-b.y)); } }; struct polygon{ int n; Point p[maxn]; struct cmp{ Point p; cmp(const Point &p0) {p=p0;} bool operator()(const Point &aa,const Point &bb){ Point a=aa,b=bb; int d=sgn((a-p)^(b-p)); if(d==0){ return sgn(a.distance(p) - b.distance(p)) < 0; } return d>0; } }; void norm(Point t){ sort(p+1,p+1+n,cmp(t)); } int relationpoint(Point q){ int cnt = 0; fo(i,0,n-1){ int j = (i+1)%n; int k = sgn((q-p[j])^(p[i]-p[j])); int u = sgn(p[i].y-q.y); int v = sgn(p[j].y-q.y); if(k>0&&u<0&&v>=0)cnt++; if(k<0&&v<0&&u>=0)cnt--; if(k==0)return 2; } return cnt != 0; } } Ps, pts, rec; int n; bool ok[maxn][maxn][maxn]; double dp[maxn][maxn],ans; void gao(int st){ memset(dp,0,sizeof(dp)); int m = n-st+1; pts.n = m; fo(i,1,m){ pts.p[i] = ps.p[st+i-1]; } pts.norm (pts.p [ 1 ]); fo(i,2,m){ (j, i + 1 , m) { if (! ok [pts.p [ 1 ] .id] [pts.p [i] .id] [pts.p [j] .id]) { dp[i][j] = 0; }else{ bool flag = true; (k, 2 , i- 1 ) { if (sgn ((pts.p [k] -pts.p [ 1 ]) ^ (pts.p [i] -pts.p [ 1 ])) == 0 ) { flag=false; break; } } if (flag) (k, 2 , i- 1 ) { if (sgn ((pts.p [i] -pts.p [k]) ^ (pts.p [j] -pts.p [i]) )> = 0 ) { dp[i][j] = max(dp[i][j],dp[k][i]); } } dp [i] [j] + = ((pts.p [i] -pts.p [ 1 ]) ^ (pts.p [j] -pts.p [ 1 ])) / 2.0 ; } years = max (years dp [i] [j]); } } } int main() { int T=read(); while(T--){ years = 0 ; n=ps.n=read(); fo(i,1,n){ ps.p[i] = Point(read(),read()); } sort (ps.p + 1 , ps.p + 1 + n); fo(i,1,n) ps.p[i].id=i; fo(i,1,n){ fo(j,i+1,n){ fo(k,j+1,n){ rec.n = 3 ; rec.p[0]=ps.p[i];rec.p[1]=ps.p[j];rec.p[2]=ps.p[k]; ok[k][i][j]=ok[k][j][i]=ok[j][i][k]=ok[j][k][i]=ok[i][k][j]=ok[i][j][k]=true; if(sgn((rec.p[1]-rec.p[0])^(rec.p[2]-rec.p[0]))==0)continue; fo(t,1,n){ if(t==i||t==j||t==k)continue; if(rec.relationpoint(ps.p[t])==1){ ok[k][i][j]=ok[k][j][i]=ok[j][i][k]=ok[j][k][i]=ok[i][k][j]=ok[i][j][k]=false; break; } } } } } fo(i,1,n){ gao (i); } printf("%.1f\n",ans); } return 0; }