# [bzoj1670][Usaco2006 Oct]Building the Moat

Description

Input

Output

Sample Input

20
2 10
3 7
22 15
12 11
20 3
28 9
1 12
9 3
14 14
25 6
8 1
25 1
28 4
24 12
4 15
13 5
26 5
21 11
24 4
1 8

Sample Output

70.87

HINT

$8\;\leq\;N\;\leq\;5000$.

Solution

#include<set>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 5005
#define eps 1e-11
using namespace std;
struct point{
int x,y;double t;
}a[N],v[N];
int n,u,vn;
inline double sqr(int k){
return (double)(k*k);
}
inline point dec(point x,point y){
return (point){x.x-y.x,x.y-y.y,0.0};
}
inline int mult(point x,point y){
return x.x*y.y-y.x*x.y;
}
inline double dis(point x,point y){
return sqrt(sqr(abs(x.x-y.x))+sqr(abs(x.y-y.y)));
}
inline bool cmp(point x,point y){
if(fabs(x.t-y.t)<eps)
return dis(x,a[1])>dis(y,a[1]);
return x.t<y.t;
}
inline void convex(){
u=1;
for(int i=2;i<=n;i++)
if((a[i].x<a[u].x)||(a[i].x==a[u].x&&a[i].y<a[u].y)) u=i;
a[0]=a[u];a[u]=a[1];a[1]=a[0];
for(int i=2;i<=n;i++)
a[i].t=atan2(a[i].y-a[1].y,a[i].x-a[1].x);
sort(a+2,a+1+n,cmp);
v[++vn]=a[1];v[++vn]=a[2];a[++n]=a[1];
for(int i=3;i<=n;i++){
if(fabs(a[i].t-a[i-1].t)<eps)
continue;
while(vn>1&&mult(dec(a[i],v[vn-1]),dec(v[vn],v[vn-1]))>0) vn--;
v[++vn]=a[i];
}
}
inline double cir(){
double ret=0;
for(int i=2;i<=vn;i++)
ret+=dis(v[i],v[i-1]);
return ret;
}
inline void init(){
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d%d",&a[i].x,&a[i].y);
convex();
printf("%.2lf\n",cir());
}
int main(){
freopen("convex.in","r",stdin);
freopen("convex.out","w",stdout);
init();
fclose(stdin);
fclose(stdout);
return 0;
}
posted @ 2017-01-05 00:25  Aireen_Ye  阅读(185)  评论(0编辑  收藏  举报
Der Erfolg kommt nicht zu dir, du musst auf den Erfolg zugehen.