凸包模板
在此献上凸包模板一份,来自https://blog.csdn.net/yang_deyuan/article/details/78863424
//凸包模板,以POJ3348为例
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
typedef pair<int, int> PII;
typedef long long LL;
typedef unsigned long long ULL;
const double eps = 1e-8;
int sgn(double x) {
if(fabs(x) < eps)return 0;
if(x < 0)return -1;
else return 1;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y) {
x = _x;y = _y;
}
Point operator -(const Point &b)const {
return Point(x - b.x,y - b.y);
}
//叉积
double operator ^(const Point &b)const {
return x*b.y - y*b.x;
}
//点积
double operator *(const Point &b)const {
return x*b.x + y*b.y;
}
void input(){
scanf("%lf%lf",&x,&y);
}
};
struct Line {
Point s,e;
Line(){}
Line(Point _s,Point _e) {
s = _s; e = _e;
}
};
//*两点间距离
double dist(Point a,Point b)
{
return sqrt((a-b)*(a-b));
}
/*
* 求凸包,Graham算法
* 点的编号0~n-1
* 返回凸包结果Stack[0~top-1]为凸包的编号
*/
const int MAXN = 1010;
Point List[MAXN];
int Stack[MAXN];//用来存放凸包的点
int top;//表示凸包中点的个数
//相对于List[0]的极角排序
bool _cmp(Point p1,Point p2) {
double tmp = (p1-List[0])^(p2-List[0]);
if(sgn(tmp) > 0)
return true;
else if(sgn(tmp) == 0 && sgn(dist(p1,List[0]) - dist(p2,List[0])) <= 0)
return true;
else
return false;
}
void Graham(int n) {
Point p0;
int k = 0;
p0 = List[0];
//找最下边的一个点
for(int i = 1;i < n;i++) {
if( (p0.y > List[i].y) || (p0.y == List[i].y && p0.x > List[i].x) ) {
p0 = List[i];
k = i;
}
}
swap(List[k],List[0]);
sort(List+1,List+n,_cmp);
if(n == 1) {
top = 1;
Stack[0] = 0;
}
else {
Stack[0] = 0;
Stack[1] = 1;
top = 2;
for(int i = 2;i < n;i++) {
while(top > 1 && sgn((List[Stack[top-1]]-List[Stack[top-2]])^(List[i]-List[Stack[top-2]])) <= 0)
top--;
Stack[top++] = i;
}
}
}
int n;
int main() {
scanf("%d", &n);
for(int i = 0; i < n; i++)List[i].input();
Graham(n);
double res = 0;
for(int i = 0; i < top; i++) res += List[Stack[i]] ^ List[Stack[(i + 1) % top]];
printf("%d\n", (int)(res / 100));
return 0;
}