HDU1392 Surround the Trees

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题目链接：HDU1392

正解：凸包

解题报告：

　　凸包模板题。

//It is made by ljh2000
//有志者，事竟成，破釜沉舟，百二秦关终属楚；苦心人，天不负，卧薪尝胆，三千越甲可吞吴。
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <complex>
#include <vector>
#include <cstdio>
#include <string>
#include <bitset>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define lc root<<1
#define rc root<<1|1
#define pr pair<int,int>
#define MP make_pair
#define fr first
#define sc second
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define reg(i,x) for(int i=first[x];i;i=next[i])
using namespace std;
typedef long long LL;
typedef long double LB;
typedef complex<double> C;
const double pi = acos(-1);
const double eps = 1e-9;
const int MAXN = 1011;
int n,top,stack[MAXN];
struct node{ double x,y; }a[MAXN];
inline node operator - (const node &q,const node &qq) { return (node){q.x-qq.x,q.y-qq.y}; }
inline double cross(node q,node qq){ return q.x*qq.y-q.y*qq.x; }
inline bool cmp(node q,node qq){ if(q.x==qq.x) return q.y<qq.y; return q.x<qq.x; }
inline double sqr(LL x){ return x*x; }
inline double dis(node q,node qq){ return sqrt(sqr(q.x-qq.x)+sqr(q.y-qq.y)); }
inline int getint(){
    int w=0,q=0; char c=getchar(); while((c<'0'||c>'9') && c!='-') c=getchar();
    if(c=='-') q=1,c=getchar(); while (c>='0'&&c<='9') w=w*10+c-'0',c=getchar(); return q?-w:w;
}

inline void work(){
	while(1) {
		n=getint(); if(n==0) break;
		for(int i=1;i<=n;i++) a[i].x=getint(),a[i].y=getint();
		if(n==1) printf("%.2lf
",0.00);  
        else if(n==2) printf("%.2lf
",dis(a[1],a[2]));  
		else {
			sort(a+1,a+n+1,cmp); stack[top=1]=1;
			for(int i=2;i<=n;i++) {
				while(top>1 && cross( a[stack[top]]-a[stack[top-1]] , a[i]-a[stack[top-1]]) <= 0 ) top--;
				stack[++top]=i;
			}
			int k=top;
			for(int i=n;i>=1;i--) {
				while(top>k && cross( a[stack[top]]-a[stack[top-1]] , a[i]-a[stack[top-1]]) <= 0 ) top--;
				stack[++top]=i;
			}
			if(n!=1) top--; double ans=0;
			for(int i=1;i<=top/*!!!*/;i++) ans+=dis(a[stack[i]],a[stack[i+1]]);
			printf("%.2lf
",ans);
		}
	}
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("1392.in","r",stdin);
	freopen("1392.out","w",stdout);
#endif
    work();
    return 0;
}
//有志者，事竟成，破釜沉舟，百二秦关终属楚；苦心人，天不负，卧薪尝胆，三千越甲可吞吴。