牛客多校(2020第三场)C Operation Love

题目链接： https://ac.nowcoder.com/acm/contest/5668/C

题意：给了右手手掌的形状，然后给你20个连续的点（可顺可逆），问是左手还是右手

题解：

　　寻找最长边的那俩个点，和紧挨它的短的那条边的点，共三个点，求叉积判断左右手

 1 #include
 2 using namespace std;
 3 
 4 const int N = 30;
 5 double x[N], y[N];
 6 
 7 double cal(double x1, double y1, double x2, double y2) { //计算俩点距离的平方
 8     return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
 9 }
10 
11 int main() {
12     int t;
13     scanf("%d", & t);
14     while (t--) {
15         for (int i = 1; i <= 20; i++) {
16             scanf("%lf %lf", &x[i], &y[i]);
17         }  
18         x[0] = x[20], y[0] = y[20];
19         x[21] = x[1], y[21] = y[1];
20         x[22] = x[2], y[22] = y[2]; //若不懂这三行，且先别急，看完33-39就明白了
21 
22         double max_len = -1;
23 
24         int pos1 = 0, pos2 = 0; //存最长边10的俩个端点
25         for (int i = 1; i <= 20; i++) {
26             double temp = cal(x[i], y[i], x[i+1], y[i+1]);
27             if (max_len < temp) {
28                 max_len = temp;
29                 pos1 = i, pos2 = i+1; //记录下最长边的俩端
30             }
31         }
32 
33         int l, mid, r;
34         if (cal(x[pos1], y[pos1], x[pos1 - 1], y[pos1 -1]) < cal(x[pos2], y[pos2], x[pos2+1], y[pos2+1])) {
35             l = pos1 - 1, mid = pos1, r = pos2;
36         }
37         else {
38             l = pos2 + 1, mid = pos2, r = pos1;
39         }
40 
41         if (((x[r] - x[mid]) * (y[l] -y[mid]) - (x[l] - x[mid]) * (y[r] - y[mid])) < 0) {
42             printf("left\n");
43         }
44         else {
45             printf("right\n");
46         }
47     }
48     
49     return 0;
50 }