Wall
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3139 Accepted Submission(s): 888
Your task is to help poor Architect to save his head, by writing a program that will find the minimum possible length of the wall that he could build around the castle to satisfy King's requirements.
The task is somewhat simplified by the fact, that the King's castle has a polygonal shape and is situated on a flat ground. The Architect has already established a Cartesian coordinate system and has precisely measured the coordinates of all castle's vertices in feet.
Next N lines describe coordinates of castle's vertices in a clockwise order. Each line contains two integer numbers Xi and Yi separated by a space (-10000 <= Xi, Yi <= 10000) that represent the coordinates of ith vertex. All vertices are different and the sides of the castle do not intersect anywhere except for vertices.
This problem contains multiple test cases!
The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.
The output format consists of N output blocks. There is a blank line between output blocks.
简单凸包,题意和我的想法有点出入,既然是要求最小的长度,就不是简单的凸包可以解决的,额,忽略细节的话就是凸包模板题。
求凸包的长度+半径为L的圆的周长。
//31MS 256K 1479B G++
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 1005
#define pi 3.1415926
struct node{
double x,y;
}p[N],stack[N];
double dist(node a,node b)
{
return sqrt((a.y-b.y)*(a.y-b.y)+(a.x-b.x)*(a.x-b.x));
}
double crossprod(node a,node b,node c)
{
return ((b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y))/;
}
int cmp(const void*a,const void*b)
{
node c=*(node*)a;
node d=*(node*)b;
double k=crossprod(p[],c,d);
if(k< || !k && dist(p[],c)>dist(p[],d))
return ;
return -;
}
double graham(int n)
{
for(int i=;i<n;i++)
if(p[i].x<p[].x || p[i].x==p[].x && p[i].y<p[].y){
node temp=p[];
p[]=p[i];
p[i]=temp;
}
qsort(p+,n-,sizeof(p[]),cmp);
p[n]=p[];
for(int i=;i<;i++) stack[i]=p[i];
int top=;
for(int i=;i<n;i++){
while(crossprod(stack[top-],stack[top],p[i])<= && top>=)
top--;
stack[++top]=p[i];
}
double ans=dist(stack[],stack[top]);
for(int i=;i<top;i++)
ans+=dist(stack[i],stack[i+]);
return ans;
}
int main(void)
{
int t,n;
double l;
scanf("%d",&t);
while(t--)
{
scanf("%d%lf",&n,&l);
for(int i=;i<n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
double ans=*l*pi;
ans+=graham(n);
printf("%.0lf\n",ans);
if(t) printf("\n");
}
return ;
}