- POJ 1113 - Wall
- Time: 1000MS
- Memory: 10000K
- 难度: 初级
- 分类: 凸包
问题描述
给定多边形城堡的n个顶点,绕城堡外面建一个围墙,围住所有点,并且墙与所有点的距离至少为L,求这个墙最小的长度。
解题思路
推导公式(1):城堡围墙长度最小值 = 城堡顶点坐标构成的散点集的凸包总边长 + 半径为L的圆周长
由于数据规模较大,必须用GrahamScan Algorithm构造凸包(详细的算法可以参考我的 [POJ2187](.//Memory Time
//244K 63MS
#include
#include
#include
using namespace std;
const int inf=10001;
const double pi=3.141592654;
typedef class
{
public:
int x,y;
}point;
/AB距离平方/
int distsquare(point A,point B)
{
return (B.x-A.x)(B.x-A.x)+(B.y-A.y)(B.y-A.y);
}
/AB距离/
double distant(point A,point B)
{
return sqrt((double)((B.x-A.x)(B.x-A.x)+(B.y-A.y)(B.y-A.y)));
}
/叉积计算/
int det(int x1,int y1,int x2,int y2)
{
return x1y2-x2y1;
}
int cross(point A,point B,point C,point D)
{
return det(B.x-A.x,B.y-A.y,D.x-C.x,D.y-C.y);
}
/快排判断规则/
point* s;
int cmp(const void* pa,const void* pb)
{
point* a=(point*)pa;
point* b=(point*)pb;
int temp=cross(*s,*a,*s,*b);
if(temp>0)
return -1;
else if(temp==0)
return distsquare(*s,*b)-distsquare(*s,*a);
else
return 1;
}
int main(int i,int j)
{
int N,L;
while(cin>>N>>L)
{
/Input/
point* node=new point[N+1];
int min_x=inf;
int fi;
for(i=1;i<=N;i++)
{
cin>>node[i].x>>node[i].y;
if(min_x > node[i].x)
{
min_x = node[i].x;
fi=i;
}
else if(min_x == node[i].x)
if(node[fi].y > node[i].y)
fi=i;
}
/*Quicksort the Vertex*/
node[0]=node[N];
node[N]=node[fi];
node[fi]=node[0];
s=&node[N];
qsort(node+1,N,sizeof(point),cmp);
/*Structure Con-bag*/
int* bag=new int[N+2];
bag[1]=N;
bag[2]=1;
int pb=2;
for(i=2;i<=N;)
if(cross(node[ bag[pb-1] ],node[ bag[pb] ],node[ bag[pb] ],node[i]) >= 0)
bag[++pb]=i++;
else
pb--;
/*Compute Min-length*/
double minlen=0;
for(i=1;i<pb;i++)
minlen+=distant(node[ bag[i] ],node[ bag[i+1] ]);
minlen+=2*pi*L;
cout<<fixed<<setprecision(0)<<minlen<<endl;
delete node;
delete bag;
}
return 0;
}
------
## 相关资料
- [北大 ACM - POJ 试题分类](https://exp-blog.com/algorithm/poj-shi-ti-fen-lei/)
- [北大 POJ 题库(官网在线)](http://poj.org/)
- [北大 POJ 题库(离线版)](https://github.com/lyy289065406/POJ-Solving-Reports/doc/POJ%E7%A6%BB%E7%BA%BF%E7%89%88%E9%A2%98%E7%9B%AE.chm)
- [POJ封面书《程序设计导引及在线实践》](https://github.com/lyy289065406/POJ-Solving-Reports/doc/程序设计导引及在线实践.pdf)
- [ACM 资料](https://lyy289065406.github.io/articles/tags/ACM/)