hdoj Surround the Trees 1392 （凸包）

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9158    Accepted Submission(s): 3508


[align=left]Problem Description[/align]
There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?

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.

[align=left]Input[/align]
The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer
is less than 32767. Each pair is separated by blank.

Zero at line for number of trees terminates the input for your program.

[align=left]Output[/align]
The minimal length of the rope. The precision should be 10^-2.

[align=left]Sample Input[/align]

9
12 7
24 9
30 5
41 9
80 7
50 87
22 9
45 1
50 7
0

[align=left]Sample Output[/align]

243.06

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct zz
{
int x;
int y;
}q[1010],p[1010];
int cmp(zz a,zz b)
{
if(a.x==b.x)
return a.y<b.y;
return a.x<b.x;
}
int judge(zz a,zz b,zz c)
{
return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
int main()
{
int t,i,j;
int n,r;
while(scanf("%d",&n),n)
{
for(i=0;i<n;i++)
scanf("%d%d",&q[i].x,&q[i].y);
sort(q,q+n,cmp);
int k1=0;
for(i=0;i<n;i++)
{
while(k1>1&&judge(p[k1-2],p[k1-1],q[i])<=0)
k1--;
p[k1++]=q[i];
}
int k2=k1;
for(i=n-1;i>=0;i--)
{
while(k1>k2&&judge(p[k1-2],p[k1-1],q[i])<=0)
k1--;
p[k1++]=q[i];
}
k1--;
double sum=0;
for(i=0,j=1;i<k1;i++,j++)
{
sum+=sqrt((p[i].x-p[j].x)*(p[i].x-p[j].x)+(p[i].y-p[j].y)*(p[i].y-p[j].y));
}
if(n>2)
printf("%.2lf\n",sum);
else
printf("%.2lf\n",sum/2);//如果点的个数为二，只有一条线，不能组成闭合图，所以在计算时多算了一遍，在这块要除2.
}
return 0;
}