HDOJ题目1392Surround the Trees(数学几何,凸包模板)
2014-09-04 23:18
711 查看
Total Submission(s): 7644 Accepted Submission(s): 2921
[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
[align=left]Source[/align]
Asia 1997, Shanghai (Mainland
China)
[align=left]Recommend[/align]
Ignatius.L | We have carefully selected several similar problems for you: 1115 2150 1348 1147 1558
凸包:http://www.cnblogs.com/jbelial/archive/2011/08/05/2128625.html
ac代码
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
struct s
{
double x,y;
}p[101],temp;
int stack[101];
int top;
double muti(struct s pi,struct s pj,struct s pk)
{
return (pj.y-pk.y)*(pi.x-pk.x)-(pj.x-pk.x)*(pi.y-pk.y);
}
int cmp(const void *a,const void *b)
{
struct s *c=(struct s *)a;
struct s *d=(struct s *)b;
double k=muti(*c,*d,p[0]);
if(k<=0)
return 1;
else
return -1;
}
double dis(struct s a,struct s b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF,n)
{
int i,j,k,u=0;
double sum=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
if(n<2)
{
printf("0.00\n");
continue;
}
if(n==2)
{
printf("%.2lf\n",dis(p[0],p[1]));
continue;
}
for(i=0;i<n;i++)
{
if(p[i].y<p[u].y||((p[i].y==p[u].y)&&p[i].x<p[u].x))
{
/*temp=p[i];
p[i]=p[0];
p[0]=temp;*/
u=i;
}
}
temp=p[u];
p[u]=p[0];
p[0]=temp;
qsort(p+1,n-1,sizeof(p[1]),cmp);
for(i=0;i<=2;i++)
stack[i]=i;
top=2;
for(i=3;i<n;i++)
{
while(muti(p[stack[top]],p[i],p[stack[top-1]])<=0)
top--;
stack[++top]=i;
}
for(i=1;i<=top;i++)
{
sum+=dis(p[stack[i]],p[stack[i-1]]);
}
sum+=dis(p[stack[0]],p[stack[top]]);
printf("%.2lf\n",sum);
}
}
Surround the Trees
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7644 Accepted Submission(s): 2921
[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
[align=left]Source[/align]
Asia 1997, Shanghai (Mainland
China)
[align=left]Recommend[/align]
Ignatius.L | We have carefully selected several similar problems for you: 1115 2150 1348 1147 1558
凸包:http://www.cnblogs.com/jbelial/archive/2011/08/05/2128625.html
ac代码
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
struct s
{
double x,y;
}p[101],temp;
int stack[101];
int top;
double muti(struct s pi,struct s pj,struct s pk)
{
return (pj.y-pk.y)*(pi.x-pk.x)-(pj.x-pk.x)*(pi.y-pk.y);
}
int cmp(const void *a,const void *b)
{
struct s *c=(struct s *)a;
struct s *d=(struct s *)b;
double k=muti(*c,*d,p[0]);
if(k<=0)
return 1;
else
return -1;
}
double dis(struct s a,struct s b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF,n)
{
int i,j,k,u=0;
double sum=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
}
if(n<2)
{
printf("0.00\n");
continue;
}
if(n==2)
{
printf("%.2lf\n",dis(p[0],p[1]));
continue;
}
for(i=0;i<n;i++)
{
if(p[i].y<p[u].y||((p[i].y==p[u].y)&&p[i].x<p[u].x))
{
/*temp=p[i];
p[i]=p[0];
p[0]=temp;*/
u=i;
}
}
temp=p[u];
p[u]=p[0];
p[0]=temp;
qsort(p+1,n-1,sizeof(p[1]),cmp);
for(i=0;i<=2;i++)
stack[i]=i;
top=2;
for(i=3;i<n;i++)
{
while(muti(p[stack[top]],p[i],p[stack[top-1]])<=0)
top--;
stack[++top]=i;
}
for(i=1;i<=top;i++)
{
sum+=dis(p[stack[i]],p[stack[i-1]]);
}
sum+=dis(p[stack[0]],p[stack[top]]);
printf("%.2lf\n",sum);
}
}
相关文章推荐
- hdu 1392 Surround the Trees 凸包模板题
- hdu 1392 Surround the Trees(凸包模板题)
- hdu 1392 Surround the Trees(凸包模板)
- HDU 1392 Surround the Trees(凸包模板)
- 杭电OJ(HDOJ)1392题:Surround the Trees(凸包问题)
- hdoj Surround the Trees 1392 (凸包)
- HDOJ 1392 Surround the Trees【凸包周长 Graham扫描】
- HDOJ_1392 Surround The Trees (凸包模版)
- hdu 1392 Surround the Trees 凸包模板
- hdoj 1392 Surround the Trees 【求凸包周长】
- HDU 1392 Surround the Trees(凸包模板)
- HDU 1392 Surround the Trees(计算几何,求凸包周长)
- 计算几何(凸包模板):HDU 1392 Surround the Trees
- hdu 1392:Surround the Trees(计算几何,求凸包周长)
- HDU 1392 Surround the Trees 凸包的周长
- 【计算几何初步-凸包-Jarvis步进法。】【HDU1392】Surround the Trees
- HDU 1392 Surround the Trees 构造凸包
- HDOJ1392 Surround the Trees
- hdu 1392 Surround the Trees(凸包模版)
- hdu 1392 Surround the Trees( 凸包问题)