●POJ 1113 Wall

题链：

题解：

计算几何，凸包

题意：修一圈围墙把给出的点包围起来，且被包围的点距离围墙的距离不能小于L，求围墙最短为多少。

答案其实就是等于N个点的凸包的周长+半径为L的圆的周长。

代码：

#include<cmath>

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define MAXN 1050

using namespace std;

const double eps=1e-8,Pi=acos(-1.0);

int sign(double x){

	if(fabs(x)<=eps) return 0;

	return x<0?-1:1;

}

struct Point{

	double x,y;

	Point(double _x=0,double _y=0):x(_x),y(_y){}

	void Read(){scanf("%lf%lf",&x,&y);}

};

typedef Point Vector;

bool operator < (const Point A,const Point B){return sign(A.x-B.x)<0||(sign(A.x-B.x)==0&&sign(A.y-B.y)<0);}

bool operator == (const Point A,const Point B){return sign(A.x-B.x)==0&&sign(A.y-B.y)==0;}

Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}

double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}

double operator * (Vector A,Vector B){return A.x*B.x+A.y*B.y;}

Point D[MAXN],H[MAXN];

int N,L;

int Andrew(int dnt){

	int hnt=0,k=0;

	sort(D+1,D+dnt+1);

	dnt=unique(D+1,D+dnt+1)-D-1;

	for(int i=1;i<=dnt;i++){

		while(hnt>1&&(sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1])))<=0) hnt--;

		H[++hnt]=D[i];

	} k=hnt;

	for(int i=dnt-1;i>=1;i--){

		while(hnt>k&&(sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1])))<=0) hnt--;

		H[++hnt]=D[i];

	}

	return hnt;

}

double GL(Vector A){//Get_Length

	return sqrt(A*A);

}

double GPC(int hnt){//Get_Polygon_Circumference

	double C=0;

	for(int i=1;i<hnt;i++) C+=GL(H[i]-H[i+1]);

	return C;

}

int main(){

	while(~scanf("%d%d",&N,&L)){

		for(int i=1;i<=N;i++) D[i].Read();

		N=Andrew(N);

		printf("%.0f\n",GPC(N)+Pi*2*L);

	}

	return 0;

}

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