题意:给你一堆点,求一个最大面积的空凸包,里面没有点.
题解:红书板子,照抄完事,因为题目给的都是整点,所以最后答案一定是.5或者.0结尾,不用对答案多做处理
1 #include<bits/stdc++.h> 2 #define N 55 3 using namespace std; 4 struct rec 5 { 6 double x,y; 7 }; 8 rec operator -(rec a,rec b) 9 { 10 rec c; 11 c.x=a.x-b.x; 12 c.y=a.y-b.y; 13 return c; 14 } 15 double sqr(double a) 16 { 17 return a*a; 18 } 19 int sign(double a) 20 { 21 if (fabs(a) <= 1e-6) return 0; 22 return a<0?-1 :1; 23 } 24 bool operator <(rec a,rec b) 25 { 26 return sign(b.y-a.y)>0 || sign(b.y-a.y)==0 && sign(b.x-a.x)>0; 27 } 28 double max(double a,double b) 29 { 30 return a>b ?a:b; 31 } 32 double length(rec a) 33 { 34 return sqrt(sqr(a.x)+sqr(a.y)); 35 } 36 double cross(rec a,rec b) 37 { 38 return a.x*b.y-a.y*b.x; 39 } 40 rec dot[N],lis[N]; 41 double opt[N][N]; 42 int seq[N],n,len; 43 double ans; 44 bool Compare(rec a,rec b) 45 { 46 int temp=sign(cross(a,b)); 47 if (temp!=0) return temp>0; 48 temp=sign(length(b)-length(a)); 49 return temp>0; 50 } 51 void solve(int vv) 52 { 53 int t,i,j,_len; 54 for (i=len=0;i<n;i++) 55 if (dot[vv]<dot[i]) lis[len++]=dot[i]-dot[vv]; 56 for (int i=0;i<len;i++) 57 for (int j=0;j<len;j++) 58 opt[i][j]=0; 59 sort(lis,lis+len,Compare); 60 double v; 61 for (t=1;t<len;t++) 62 { 63 _len=0; 64 for (i=t-1;i>=0 && sign(cross(lis[t],lis[i])) ==0 ;i--); 65 while (i>=0) 66 { 67 v=cross(lis[i],lis[t])/2; 68 seq[_len++]=i; 69 for (j=i-1; j>=0 && sign(cross(lis[i]-lis[t], lis[j]-lis[t])) >0 ;j--); 70 if (j>=0) v+=opt[i][j]; 71 ans=max(ans,v); 72 opt[t][i]=v; 73 i=j; 74 } 75 for (i = _len-2;i>=0;i--) 76 opt[t][seq[i]]=max(opt[t][seq[i]],opt[t][seq[i+1]]); 77 } 78 } 79 int T; 80 int main() 81 { 82 scanf("%d",&T); 83 while (T--) 84 { 85 scanf("%d",&n); 86 for (int i=0;i<n;i++) scanf("%lf%lf",&dot[i].x,&dot[i].y); 87 ans=0; 88 for (int i=0;i<n;i++) solve(i); 89 printf("%.1lf ",ans); 90 } 91 return 0; 92 }