USACO 5.1 Fencing the Cows

Farmer John wishes to build a fence to contain his cows, but he's a bit short on cash right. Any fence he builds must contain all of the favorite grazing spots for his cows. Given the location of these spots, determine the length of the shortest fence which encloses them.

PROGRAM NAME: fc

INPUT FORMAT

The first line of the input file contains one integer, N. N (0 <= N <= 10,000) is the number of grazing spots that Farmer john wishes to enclose. The next N line consists of two real numbers, Xi and Yi, corresponding to the location of the grazing spots in the plane (-1,000,000 <= Xi,Yi <= 1,000,000). The numbers will be in decimal format.

SAMPLE INPUT (file fc.in)

4
4 8
4 12
5 9.3
7 8

OUTPUT FORMAT

The output should consists of one real number, the length of fence required. The output should be accurate to two decimal places.

SAMPLE OUTPUT (file fc.out)

12.00

————————————————————————————————题解

一道凸包求周长的题，卷包裹算法，如果栈中的后两个点组成的向量和新加入的点和最后点的向量是顺时针旋转，那么就不符合凸包的性质，弹出直到符合性质即可

1 /*
2 ID: ivorysi
3 LANG: C++
4 TASK: fc
5 */
6 #include <iostream>
7 #include <cstdio>
8 #include <cstring>
9 #include <algorithm>
10 #include <queue>
11 #include <set>
12 #include <vector>
13 #include <string.h>
14 #include <cmath>
15 #define siji(i,x,y) for(int i=(x);i<=(y);++i)
16 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j)
17 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)
18 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j)
19 #define inf 0x7fffffff
20 #define ivorysi
21 #define mo 97797977
22 #define hash 974711
23 #define base 47
24 #define pss pair<string,string>
25 #define MAXN 5000
26 #define fi first
27 #define se second
28 #define pii pair<int,int>
29 #define esp 1e-8
30 typedef long long ll;
31 using namespace std;
32 struct vec {
33     double x,y;
34     vec(double x=0.0,double y=0.0) :x(x),y(y) {}
35     vec operator - (const vec &b) const{
36         return vec(x-b.x,y-b.y);
37     }
38     double norm() const{
39         return x*x+y*y;
40     }
41 }a[10005];
42 int n;
43 bool dcmp(double a,double b=0.0) {
44     return fabs(a-b) < esp ? 1 :0 ;
45 }
46 double cross(vec a,vec b) {
47     return  a.x*b.y-a.y*b.x;
48 }
49 inline bool cmp(const vec &c,const vec &d) {
50     vec t1= c-a[1],t2=d-a[1];
51     double t=cross(t1,t2);
52     if(!dcmp(t)) return t>0;//逆时针为正，顺时针为负
53     else return c.norm()<d.norm();
54 }
55 double o(double a){
56     return a*a;
57 }
58 struct poly {
59     vector<vec> poi;
60     double peri() {
61         double res=0.0;
62         siji(i,1,poi.size()-1) {
63             res+=sqrt(o(poi[i].x-poi[i-1].x)+o(poi[i].y-poi[i-1].y));
64         }
65         res+=sqrt(o(poi[0].x-poi[poi.size()-1].x)+o(poi[0].y-poi[poi.size()-1].y));
66         return res;
67     }
68     void convex() {
69         int id=1;
70         siji(i,2,n) {
71             if(a[i].x<a[id].x || (a[i].x==a[id].x && a[i].y<a[id].y)) {
72                 id=i;
73             }
74         }
75         if(id!=1) swap(a[id],a[1]);
76         sort(a+2,a+1+n,&cmp);
77         poi.push_back(a[1]);
78         siji(i,2,n) {
79             while(poi.size() >=2 &&
80                 cross(poi[poi.size()-1]-poi[poi.size()-2],a[i]-poi[poi.size()-1])<=0)
81                 poi.pop_back();
82             poi.push_back(a[i]);
83         }
84     }
85 }tw;
86
87 void solve() {
88     scanf("%d",&n);
89     siji(i,1,n) {
90         scanf("%lf%lf",&a[i].x,&a[i].y);
91     }
92     tw.convex();
93     printf("%.2lf\n",tw.peri());
94 }
95 int main(int argc, char const *argv[])
96 {
97 #ifdef ivorysi
98     freopen("fc.in","r",stdin);
99     freopen("fc.out","w",stdout);
100 #else
101     freopen("f1.in","r",stdin);
102 #endif
103     solve();
104     return 0;
105 }