HDUJ 1392 Surround the Trees 凸包
Surround the Trees
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7203 Accepted Submission(s): 2752
The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.
There are no more than 100 trees.
is less than 32767. Each pair is separated by blank.
Zero at line for number of trees terminates the input for your program.
9
12 7
24 9
30 5
41 9
80 7
50 87
22 9
45 1
50 7
0
243.06
水平序的Andrew算法:
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std; struct node
{
double x,y;
}a[105],b[105]; double cmp(node n,node m) //先比較X坐标,在比較Y坐标(从小到大)
{
if(n.x != m.x)
return n.x < m.x;
else
return n.y < m.y;
} double Cross(node a,node b,node c) //计算叉积大小
{
return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
} double dis(node a,node b) //计算距离
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} int CH(node* a,int n,node* b)
{
sort(a,a+n,cmp);
int m=0,i;
for(i=0;i<n;i++) //从左往右,选下边界
{
while(m > 1 && Cross(b[m-2],b[m-1],a[i]) < 0)
m--;
b[m++]=a[i];
} int k=m;
for(i=n-2;i>=0;i--) //从右往左,选上边界
{
while(m > k && Cross(b[m-2],b[m-1],a[i]) < 0)
m--;
b[m++]=a[i];
} if(n >1) m--;
return m;
} int main()
{
int n;
while(cin>>n)
{
if(n==0) break;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b)); int i,j;
for(i=0;i<n;i++)
{
cin>>a[i].x>>a[i].y;
} // cout<<CH(a,n,b)<<endl; //输出所选点的总数
if(n==1)
cout<<0.00<<endl;
else if(n==2)
printf("%.2lf\n",dis(a[0],a[1]));
else
{
int m=CH(a,n,b);
double s=0;
for(i=1;i<m;i++)
s+=dis(b[i-1],b[i]);
s+=dis(b[0],b[m-1]);
printf("%.2lf\n",s);
}
// for(i=0;i<CH(a,n,b);i++) //输出所选点的坐标
// cout<<b[i].x<<" "<<b[i].y<<endl; } return 0;
}
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