10.1 cross product, the polar angle sorting, scanning seeking convex hull

Convex hull: a convex polygon with all points enclosed, this convex polygon is a convex hull

 

1. The first to introduce a mathematical tool vector cross product

 

 

 

 | C | = | A × b | = | A | | b | sin [alpha]    ([alpha] is a, the angle between the vectors b)

The | C | parallel vectors a, b consisting of the area of ​​the quadrangle

Here is the cross product of two vectors is determined relative positional relationship of (useful!)

 

 

 The axb <0 (a counterclockwise direction in the b), bxa> 0 (b in a clockwise direction)

//求叉积
struct node{
    double x ,y;
    node operator -( const node & s ){
        return {x-s.x , y-s.y};
    }
}p[N] ,s[N];

inline double X( node a ,node b ){
    return a.x*b.y - b.x*a.y;
}

 

2. Graham scan method convex hull seek

1) identify all points in the lower left corner as the point most pole

    //找左下边界点
     int k = 1;
     rep( i ,2 ,n ){
         if( p[i].y < p[k].y || p[i].y == p[k].y && p[i].x < p[k].x )
            k = i;
     }
     swap( p[1] ,p[k] );

 

2) a polar angle using the cross product for ordering

// polar angle comparison 
BOOL CMP (& Node A, Node & B) {
         Double X = X-(AP [ . 1 ], BP [ . 1 ]);
         // cross product of vectors determined positional relationship 
        IF (X> 0 ) return  . 1 ;
         IF (X == 0 && DIS (A, P [ . 1 ]) <DIS (B, P [ . 1 ])) return  . 1 ;
         return  0 ; 
}

     // polar angle Sort 
     Sort (P + 2 , + n-P + . 1 , CMP);

 

3) deposit convex hull

s is the convex hull of the stack deposit, t is the stack

By the following relationship with the cross product of the cut-off s

 

 

 

 

     //将凸包存在s中
     s[1] = p[1];
     s[2] = p[2];
     int t = 2;
     rep( i ,3 ,n ){
         while( t >= 2 && mul( s[t-1] ,s[t] ,p[i] )<=0 )t--;
         s[++t] = p[i];
     }

 

Then the convex hull would seek out

 

Template title:

P2742 [template] a two-dimensional convex hull / [USACO5.1] ring cow

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <unordered_map>
#define mem( a ,x ) memset( a , x ,sizeof(a) )
#define rep( i ,x ,y ) for( int i = x ; i<=y ;i++ )
#define lson l ,mid ,pos<<1
#define rson mid+1 ,r ,pos<<1|1
#define Fi first
#define Se second

using namespace std;
typedef long long ll ;
typedef pair<int ,int> pii;
typedef pair<ll ,int> pli;
const ll inf = 0x3f3f3f3f;
const int N = 1e5+5;
const ll mod = 1e9+7;

int n ,m;
//求叉积
struct node{
    double x ,y;
    node operator -( const node & s ){
        return {x-s.x , y-s.y};
    }
}p[N] ,s[N];

inline double X( node a ,node b ){
    return a.x*b.y - b.x*a.y;
}

inline double dis( node a ,node b ){
    return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

inline double mul( node a ,node b ,node c ){
    return X(b-a ,c-a);
}
//极角比较
bool cmp (  node &a , node &b ){
        double x = X(a-p[1], BP [ . 1 ]);
         // cross product of vectors determined positional relationship 
        IF (X> 0 ) return  . 1 ;
         IF (X == 0 && DIS (A, P [ . 1 ]) <DIS (B, P [ . 1 ] )) return  . 1 ;
         return  0 ; 
} 

// Graham scan 

int Graham () {
     // find the lower left boundary point 
     int K = . 1 ; 
     REP (I, 2 , n-) {
          IF (P [I] .y <P [K ] .y || P [I] .y == P [K] .y && P [I] .x < P [K] .x) 
            K= I; 
     } 
     the swap (P [ . 1 ], P [K]);
      // polar angle Sort 
     Sort (P + 2 , P + n-+ . 1 , CMP);
      // convex hull is present in s 
     s [ . 1 ] = P [ . 1 ]; 
     S [ 2 ] = P [ 2 ];
      int T = 2 ; 
     REP (I, . 3 , n-) {
          the while (T> = 2 && MUL (S [T- . 1 ], S [T], P [ I]) <= 0 ) T-- ; 
         S [ ++ T] = P [I]; 
     }

     return t;
}

int main( ){
    scanf("%d" ,&n);
    rep( i ,1 ,n ){
        scanf("%lf%lf" ,&p[i].x ,&p[i].y );
    }
    int sz = graham( );
    
    double ans = dis(s[1] ,s[sz]);
    rep( i ,1 ,sz-1 )ans += dis( s[i] ,s[i+1] ) ;

    printf("%.2f" ,ans);
    return 0;
}