UVA 11928 The Busy Dog

题意：一只狗被拴在杆子上，从起点开始按直线依次跑到给出的点最后回到起点问绕杆子几圈，逆时针为正，顺时针为负，撞到杆子输出Ouch!。

解法：用叉积判断方向，用余弦定理求出以杆子为顶点的角，加和除以2π，最后的答案处理因为精度问题wa了一篇orz……

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long
using namespace std;
const double eps = 1e-8;
struct node
{
    double x, y;
} point[10005];
double cross(node p1, node p2, node p3)//叉积
{
    return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y);
}
double len(node p1, node p2)//两点长度的平方
{
    return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y);
}
double cal(node p1, node p2, node p3)//余弦定理
{
    double a = len(p2, p3), b = len(p1, p3), c = len(p1, p2);
    return acos((b + c - a) / 2.0 / sqrt(b * c));
}
int line(node p1, node p2, node p3)//判断叉积为0时杆是否在两点中间
{
    return p3.x >= min(p1.x, p2.x) && p3.x <= max(p1.x, p2.x) && p3.y >= min(p1.y, p2.y) && p3.y <= max(p1.y, p2.y);
}
int main()
{
    int n;
    while(~scanf("%d", &n) && n)
    {
        node pole;
        scanf("%lf%lf", &pole.x, &pole.y);
        for(int i = 0; i < n; i++)
            scanf("%lf%lf", &point[i].x, &point[i].y);
        double degree = 0.0;
        int ans = 0;
        for(int i = 0; i < n - 1; i++)
        {
            double flag = cross(pole, point[i], point[i + 1]);
            double res = cal(pole, point[i], point[i + 1]);
            if((fabs(flag) <= eps) && line(point[i], point[i + 1], pole))
            {
                ans = 1;
                break;
            }
            else if(flag > eps)
                degree += res;
            else if(flag < -eps)
                degree -= res;
            else
            {
                if(line(point[i], point[i + 1], pole))
                {
                    ans = 1;
                    break;
                }
            }
        }
        double flag = cross(pole, point[n - 1], point[0]);
        double res = cal(pole, point[n - 1], point[0]);
        if((fabs(flag) <= eps) && line(point[n - 1], point[0], pole))
            ans = 1;
        else if(flag > eps)
            degree += res;
        else if(flag < -eps)
            degree -= res;
        else if(line(point[n - 1], point[0], pole))
            ans = 1;
        if(ans)
        {
            printf("Ouch!\n");
            continue;
        }
        degree = degree * 0.5 / acos(-1.0);
        char x[10];
        sprintf(x, "%.0lf", degree);
        sscanf(x, "%d", &ans);
        if(ans > 0)
            printf("+");
        printf("%d\n", ans);
    }
    return 0;
}