【模板】旋转卡壳求 面积最大的三角形 poj2079
题目链接：https://vjudge.net/problem/POJ-2079
graham跑的巨慢，Andrew跑的巨快。还好写。
有两种写法。
旋转卡壳枚举三个点的（94ms）
1 /************************************************************************* 2 > File Name: poj2079.cpp 3 # File Name: poj2079.cpp 4 # Author : xiaobuxie 5 # QQ : 760427180 6 # Email:760427180@qq.com 7 # Created Time: 2019年10月12日 星期六 17时55分28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 const int N = 5e4+9; 29 struct Point{ 30 double x,y; 31 bool operator < (const Point& b)const{ 32 return x<b.x || (x==b.x && y<b.y); 33 } 34 Point operator - (const Point& b)const{ 35 return (Point){x-b.x,y-b.y}; 36 } 37 double operator ^ (const Point& b)const{ 38 return x*b.y - b.x*y; 39 } 40 }p[N],ch[N]; 41 int n; 42 inline void init(){ 43 for(int i =0 ;i<n;++i) scanf("%lf %lf",&p[i].x,&p[i].y); 44 } 45 inline int Andrew(){ 46 sort(p,p+n); 47 int m = 0; 48 for(int i = 0; i < n;++i){ 49 while( m>1 && ((p[i] - ch[m-2]) ^ (ch[m-1] - ch[m-2])) >=0) --m; 50 ch[m++] = p[i]; 51 } 52 int k = m; 53 for(int i = n-2;i >= 0;--i){ 54 while( m>k && ((p[i] - ch[m-2]) ^ (ch[m-1] - ch[m-2])) >=0) --m; 55 ch[m++] = p[i]; 56 } 57 if(n>1) --m; 58 return m; 59 } 60 inline void rotate(){ 61 double ans =0; 62 int m = Andrew(); 63 int a = 1, b = 2; 64 for(int i = 0; i < m ;++i){ 65 while( ((ch[a] - ch[i]) ^ (ch[(b+1)%m] - ch[i])) > ((ch[a] - ch[i]) ^ (ch[b] - ch[i]))){ 66 b = (b+1)%m; 67 } 68 ans = max(ans,( (ch[a] - ch[i]) ^ (ch[b] - ch[i]) )); 69 70 while( ((ch[(a+1)%m] - ch[i]) ^ (ch[b] - ch[i])) > ((ch[a] - ch[i]) ^ (ch[b] - ch[i]))){ 71 a = (a+1)%m; 72 } 73 ans = max(ans,( (ch[a] - ch[i]) ^ (ch[b] - ch[i]) )); 74 } 75 ans /= 2.0; 76 printf("%.2f\n",ans); 77 } 78 int main(){ 79 while(~scanf("%d",&n) && n!=-1){ 80 init(); 81 rotate(); 82 } 83 return 0; 84 }
枚举两点之间差，再枚举第三个点的（391ms）
1 /************************************************************************* 2 > File Name: poj2079.cpp 3 # File Name: poj2079.cpp 4 # Author : xiaobuxie 5 # QQ : 760427180 6 # Email:760427180@qq.com 7 # Created Time: 2019年10月12日 星期六 17时55分28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 const int N = 5e4+9; 29 struct Point{ 30 double x,y; 31 bool operator < (const Point& b)const{ 32 return x<b.x || (x==b.x && y<b.y); 33 } 34 Point operator - (const Point& b)const{ 35 return (Point){x-b.x,y-b.y}; 36 } 37 double operator ^ (const Point& b)const{ 38 return x*b.y - b.x*y; 39 } 40 }p[N],ch[N]; 41 int n; 42 inline void init(){ 43 for(int i =0 ;i<n;++i) scanf("%lf %lf",&p[i].x,&p[i].y); 44 } 45 inline int Andrew(){ 46 sort(p,p+n); 47 int m = 0; 48 for(int i = 0; i < n;++i){ 49 while( m>1 && ((p[i] - ch[m-2]) ^ (ch[m-1] - ch[m-2])) >=0) --m; 50 ch[m++] = p[i]; 51 } 52 int k = m; 53 for(int i = n-2;i >= 0;--i){ 54 while( m>k && ((p[i] - ch[m-2]) ^ (ch[m-1] - ch[m-2])) >=0) --m; 55 ch[m++] = p[i]; 56 } 57 if(n>1) --m; 58 return m; 59 } 60 inline void rotate(){ 61 double ans =0; 62 int m = Andrew(); 63 ch[m] = ch[0]; 64 for(int ad = 1; ad <= m/2;++ad){ 65 int now = (ad+1) % m; 66 for(int i = 0; i < m;++i){ 67 int j = (i+ad) % m; 68 while( ((ch[j] - ch[i]) ^ (ch[now] - ch[i])) < ((ch[j] - ch[i]) ^ (ch[now+1] - ch[i]))){ 69 ++now; 70 if(now == m) now = 0; 71 } 72 ans = max(ans,( (ch[j] - ch[i]) ^ (ch[now] - ch[i]) )); 73 } 74 } 75 ans /= 2.0; 76 printf("%.2f\n",ans); 77 } 78 int main(){ 79 while(~scanf("%d",&n) && n!=-1){ 80 init(); 81 rotate(); 82 } 83 return 0; 84 }
