【模板】凸包
开始挑起计几大梁了。
先开始学凸包，找了道凸包模板题。 poj1113
模板参考了kuangbin的。
1 /************************************************************************* 2 > File Name: hdu1348.cpp 3 # File Name: hdu1348.cpp 4 # Author : xiaobuxie 5 # QQ : 760427180 6 # Email:760427180@qq.com 7 # Created Time: 2019年09月02日 星期一 20时47分27秒 8 ************************************************************************/ 9 10 #include<iostream> 11 #include<cstdio> 12 #include<map> 13 #include<cmath> 14 #include<cstring> 15 #include<set> 16 #include<queue> 17 #include<vector> 18 #include<algorithm> 19 using namespace std; 20 typedef long long ll; 21 #define inf 0x3f3f3f3f 22 #define pq priority_queue<int,vector<int>,greater<int> > 23 ll gcd(ll a,ll b){ 24 if(a<b) return gcd(b,a); 25 return b==0?a:gcd(b,a%b); 26 } 27 28 29 const double pi=acos(-1.0); 30 const int N=1000+9; 31 struct point{ 32 int x,y; 33 }a[N]; 34 point p0; 35 int sta[N]; 36 int top=0; 37 int cross(point p0,point p1,point p2){ 38 return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x); 39 } 40 double dis(point p1,point p2){ 41 return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y)); 42 } 43 bool cmp(point p1,point p2){ 44 if(atan2(p1.y-p0.y,p1.x-p0.x)!=atan2(p2.y-p0.y,p2.x-p0.x)){ 45 return (atan2(p1.y-p0.y,p1.x-p0.x))<(atan2(p2.y-p0.y,p2.x-p0.x)); 46 } 47 return p1.x<p2.x; 48 } 49 void init(int n){ 50 int i,k; 51 k=0; 52 scanf("%d%d",&a[0].x,&a[0].y); 53 p0=a[0]; 54 for(i=1;i<n;++i){ 55 scanf("%d %d",&a[i].x,&a[i].y); 56 if( (p0.y > a[i].y) || ((p0.y == a[i].y) && (p0.x > a[i].x))){ 57 p0=a[i]; 58 k=i; 59 } 60 } 61 a[k]=a[0]; 62 a[0]=p0; 63 sort(a+1,a+n,cmp); 64 } 65 void graham(int n){ 66 if(n==1){ 67 top=0; 68 sta[0]=0; 69 } 70 if(n==2){ 71 top=1; 72 sta[0]=0; 73 sta[1]=1; 74 } 75 if(n>2){ 76 sta[0]=0; sta[1]=1; top=1; 77 for(int i=2;i<n;++i){ 78 while(top>0 && cross(a[sta[top-1]],a[sta[top]],a[i])<=0) top--; 79 ++top; 80 sta[top]=i; 81 } 82 } 83 } 84 int main(){ 85 int n,l; 86 while(~scanf("%d %d",&n,&l)){ 87 init(n); 88 graham(n); 89 double ans=2*pi*l; 90 for(int i=0;i<top;++i) ans+=dis(a[sta[i]],a[sta[i+1]]); 91 ans+=dis(a[sta[0]],a[sta[top]]); 92 printf("%d\n",(int)(ans+0.5)); 93 } 94 return 0; 95 }
