【hdu 2108】Shape of HDU

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=2108

【题意】

【题解】



逆时针;

可以想象一下;

如果是凸多边形的话;

逆时针的相邻的两条边;

前一条和后一条(逆时针意义上的“后一条”)边所代表的向量;

如果做叉积的话;

其结果肯定是指向屏幕外面的;

如果突然变成凹的了;

再做叉乘的话;

指向是屏幕里面的了;

也就是说如果是凸多边形的话;

叉乘的结果应该都为正数;



【Number Of WA】



0



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 11000;

struct point
{
    double x,y;
};

double chaji(point a,point b)
{
    double x1 = a.x,y1 = a.y;
    double x2 = b.x,y2 = b.y;
    return x1*y2-x2*y1;
}

int n;
point a[N];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false);
    while (cin >> n)
    {
        if (n==0) break;
        rep1(i,1,n)
            cin >> a[i].x >> a[i].y;
        a[n+1]=a[1],a[n+2]=a[2];
        bool fi = false;
        rep1(i,1,n)
        {
            point v1 = point{a[i+1].x-a[i].x,a[i+1].y-a[i].y};
            point v2 = point{a[i+2].x-a[i+1].x,a[i+2].y-a[i+1].y};
            if (chaji(v1,v2)<0) fi = true;
        }
        if (fi)
            cout <<"concave"<<endl;
        else
            cout <<"convex"<<endl;
    }
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}