链接：http://cogs.pro/cogs/problem/problem.php?pid=896
题意：给出一些点，求出这些点的凸包。
几何第一题留念……
题意已经很明白了，求出这些点的凸包。题目没有什么可以说的，这里给出两种实现方式，供读者参考。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const int maxn=10010; 8 int n,s[maxn],top; 9 struct point 10 { 11 double x,y; 12 }p[maxn]; 13 double dis(point a,point b) 14 { 15 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 16 } 17 double operator *(point a,point b) 18 { 19 return a.x*b.y-b.x*a.y; 20 } 21 point operator -(point a,point b) 22 { 23 return (point){a.x-b.x,a.y-b.y}; 24 } 25 bool operator <(point a,point b) 26 { 27 double s=(a-p[1])*(b-p[1]); 28 if(!s)return dis(a,p[1])<dis(b,p[1]); 29 return s>0; 30 } 31 void graham() 32 { 33 s[++top]=1; 34 for(int i=2;i<=n;i++) 35 { 36 while(top>1&&(p[s[top]]-p[s[top-1]])*(p[i]-p[s[top]])<0)top--; 37 s[++top]=i; 38 } 39 } 40 int haha() 41 { 42 freopen("fc.in","r",stdin); 43 freopen("fc.out","w",stdout); 44 scanf("%d",&n);int k=1; 45 for(int i=1;i<=n;i++) 46 { 47 scanf("%lf%lf",&p[i].x,&p[i].y); 48 if(p[i].x<p[k].x&&p[i].y<p[k].y)k=i; 49 } 50 swap(p[k],p[1]);sort(p+2,p+n+1); 51 graham(); 52 double ans=0; 53 for(int i=2;i<=top;i++)ans+=dis(p[s[i]],p[s[i-1]]); 54 ans+=dis(p[s[top]],p[1]); 55 printf("%0.2lf ",ans); 56 } 57 int sb=haha(); 58 int main(){;}
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 const int maxn=10010; 8 const double eps=1e-8; 9 struct point 10 { 11 double x,y; 12 bool operator<(const point &b)const 13 { 14 return (x==b.x)?(y<b.y):(x<b.x); 15 } 16 friend point operator -(point a,point b) 17 { 18 return (point){a.x-b.x,a.y-b.y}; 19 } 20 }p[maxn]; 21 int n,s[maxn]; 22 #include<vector> 23 vector<point>convex; 24 bool vis[maxn]; 25 double dis(point a,point b) 26 { 27 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 28 } 29 double cross(point a,point b) 30 { 31 return a.x*b.y-b.x*a.y; 32 } 33 bool across(point a,point b,point c) 34 { 35 return cross(b-a,c-a)>eps; 36 } 37 double graham() 38 { 39 s[0]=0;double ans=0.0; 40 sort(p+1,p+n+1); 41 s[++s[0]]=1;s[++s[0]]=2; 42 for(int i=3;i<=n;i++) 43 { 44 while(s[0]>1&&!across(p[s[s[0]]],p[i],p[s[s[0]-1]]))s[0]--; 45 s[++s[0]]=i; 46 } 47 for(int i=1;i<=s[0];i++)vis[s[i]]=1,convex.push_back(p[s[i]]); 48 s[0]=0;s[++s[0]]=n;s[++s[0]]=n-1; 49 for(int i=n-2;i;i--) 50 { 51 while(s[0]>1&&across(p[s[s[0]]],p[s[s[0]-1]],p[i]))s[0]--; 52 s[++s[0]]=i; 53 } 54 for(int i=1;i<=s[0];i++) 55 if(!vis[s[i]])convex.push_back(p[s[i]]); 56 for(int i=0,s=convex.size();i<s;i++)ans+=dis(convex[i],convex[(i+1)%s]); 57 return ans; 58 } 59 int haha() 60 { 61 freopen("fc.in","r",stdin); 62 freopen("fc.out","w",stdout); 63 scanf("%d",&n); 64 for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); 65 printf("%0.2lf",graham()); 66 } 67 int sb=haha(); 68 int main(){;}