计蒜客 The 2018 ACM-ICPC Chinese Collegiate Programming Contest Rolling The Polygon

 1 include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <algorithm>
 6 #include <utility>
 7 #include <vector>
 8 #include <map>
 9 #include <queue>
10 #include <stack>
11 #include <cstdlib>
12 #include <cmath>
13 typedef long long ll;
14 #define lowbit(x) (x&(-x))
15 #define ls l,m,rt<<1
16 #define rs m+1,r,rt<<1|1
17 using namespace std;
18 #define pi acos(-1)
19 const int N=60;
20 struct Node{
21     int x,y;
22 }node[N];
23 int  main()
24 {
25     int t,n;
26     double tx,ty;
27     scanf("%d",&t);
28     for(int i=1;i<=t;i++)
29     {
30         scanf("%d",&n);
31         for(int i=0;i<n;i++){
32             scanf("%lf%lf",&node[i].x,&node[i].y);//lf
33         }
34         scanf("%lf%lf",&tx,&ty);
35         // 为了下面的第n次旋转
36         node[n].x=node[0].x,node[n].y=node[0].y;
37         node[n+1].x=node[1].x,node[n+1].y=node[1].y;        
38         double ans=0;
39         for(int i=0;i<n;i++)//要转n次
40         {
41             double a=sqrt(pow(node[i].x-node[i+1].x,2)+pow(node[i].y-node[i+1].y,2));
42             double b=sqrt(pow(node[i+1].x-node[i+2].x,2)+pow(node[i+1].y-node[i+2].y,2));
43             double c=sqrt(pow(node[i].x-node[i+2].x,2)+pow(node[i].y-node[i+2].y,2));
44             //利用余弦定理求角度
45             // pi-acos((a*a+b*b-c*c)/(2*a*b)) 为node[i+1]转过的角度，也就是(tx,ty)转过的角度
46             double l=pi-acos((a*a+b*b-c*c)/(2*a*b));//一个小数，因为最后也以小数形式输出。
47             //弧长=半径*弧度
48             double r=sqrt(pow(node[i+1].x-tx,2)+pow(node[i+1].y-ty,2));
49             ans+=l*r;
50         }
51         printf("Case #%d: %.3f\n",i,ans);
52     }
53     return 0;
54 }