1178 Bumpy Objects问题（虽然是presentation error,没办法了）

其实求解思路倒也不复杂，就是先找到所有顶点的凸包(convex hull)，然后找到这些凸包点中每条相邻点连成的直线，同时找到直线在多边形中的端点，如Object1中1,2,5,6这条直线，端点就是1,6。如果重心和这两个点的连线和直线都是锐角的话，那么这条直线就是稳定的，就可以找到题目要求的所谓最小点了。我的代码自己测试是没问题的，但怎么都Presentation Error，无语。找凸包用的是Graham's Scan方法。

[cpp]
#include <iostream>
#include <stack>
#include <vector>
#include <math.h>
#include <iomanip>
using namespace std;

typedef struct point
{
    double x;
    double y;
    int index;//记录点的index
}point;
int cal_max(int index, vector<int>* myv, vector<point>* vertices, point mass);
int find_stable(stack<int>* mys, vector<point>* vertices, point mass);
void find_first(vector<point>* vertices);
void sort(vector<point>* vertices);
bool is_left(int index, stack<int>* mys, vector<point>* vertices);
void find_convex_hull(vector<point>* vertices, stack<int>* mys);

int main()
{
    char name[20];
    vector<point> vertices;
    stack<int> mys;
    point mass;

    while(cin>>name)
    {
        vertices.clear();
        while(mys.size() > 0)
            mys.pop();
        if(strcmp(name,"#") == 0)
        {
            return 0;
        }
        cin>>mass.x>>mass.y;

        point tmp;int index = 0;
        while(cin>>tmp.x>>tmp.y)
        {
            if(tmp.x == 0 && tmp.y == 0)
            {
                break;
            }
            tmp.index = ++index;
            vertices.push_back(tmp);
        }
        find_convex_hull(&vertices, &mys);//寻找凸包
        int min = find_stable(&mys,&vertices,mass);
        cout<<setw(20)<<setiosflags(ios::left)<<name<<setw(1)<<min<<endl;
        //cout << name << ' ' << min << endl;
    }
}
//寻找凸包的Graham's Scan法
//vertices:点集， mys:凸包的stack
void find_convex_hull(vector<point>* vertices, stack<int>* mys)
{
    find_first(vertices);//找到y坐标最小的点，放在第一位
        sort(vertices);//根据和y0角度坐标排序
        mys->push(0);mys->push(1);mys->push(2);
        for(int i = 3; i < vertices->size(); i++)
        {
            int pos = mys->top();
            while(!is_left(i,mys,vertices))
            {
                mys->pop();
            }
            mys->push(i);
        }
}
//计算myv中index和index+1组成的直线的最大baseline
//myv:凸包中的点集; vertices:点集; mass:重心坐标
//返回最大的baseline
int cal_max(int index, vector<int>* myv, vector<point>* vertices, point mass)
{
    int start = index, end = index+1;
    if(end == myv->size())
        end = 0;
    start = myv->at(start);end = myv->at(end);
    point left, right;
    int max_index = 0, min_x = 65535, max_x = -65535;
    for(int i = 0; i < myv->size(); i++)
    {
        int ano = myv->at(i);
        double rate1 = (vertices->at(end).y - vertices->at(start).y)/(vertices->at(end).x - vertices->at(start).x);
        double rate2 = (vertices->at(ano).y - vertices->at(start).y)/(vertices->at(ano).x - vertices->at(start).x);
        if(vertices->at(ano).x == vertices->at(start).x)
            rate2 = (vertices->at(ano).y - vertices->at(end).y)/(vertices->at(ano).x - vertices->at(end).x);
        if(vertices->at(ano).x == vertices->at(start).x && vertices->at(end).x == vertices->at(start).x)
            rate2 = rate1 = 1;
        if(rate1 == rate2)
        {
            if(vertices->at(ano).index > max_index)
                max_index = vertices->at(ano).index;
            if(vertices->at(ano).x > max_x)
            {
                max_x = vertices->at(ano).x;
                right = vertices->at(ano);
            }
            if(vertices->at(ano).x < min_x)
     &

补充：软件开发 , C++ ,