基础题


2017-07-15 22:29:06


  •   writer:pprp
  • 评价,用到了叉乘,很麻烦,C++构造知识必须扎实
  • 题目如下:

我们用逆时针方向的顶点序列来表示,我们很想了解这块地的基本情况,现在请你编程判断HDU的用地是凸多边形还是凹多边形呢? 

Input输入包含多组测试数据,每组数据占2行,首先一行是一个整数n,表示多边形顶点的个数,然后一行是2×n个整数,表示逆时针顺序的n个顶点的坐标(xi,yi),n为0的时候结束输入。 
Output对于每个测试实例,如果地块的形状为凸多边形,请输出“convex”,否则输出”concave”,每个实例的输出占一行。 
Sample Input
4
0 0 1 0 1 1 0 1
0
Sample Output
convex


 代码如下;




#include <iostream>

using namespace std;

class point
{
public:
int x;
int y;
point(int a,int b):x(a),y(b) {}
point():x(0),y(0) {}
void change(int a,int b)
{
x = a;
y = b;
}
};

class vec
{
public:
point a;
point b;
point pos;
point nag;
vec():a(point(0,0)),
b(point(0,0)),pos(),nag() {}
vec(point &aa,point&bb):
a(aa),b(bb),pos(b.x-a.x,b.y-a.y),nag(a.x-b.x,a.y-b.y) {}
void change(point & aa,point &bb)
{
a.x = aa.x;
a.y = aa.y;
b.x = bb.x;
b.y = bb.y;

pos.x = b.x-a.x;
pos.y = b.y-a.y;
nag.x = a.x-b.x;
nag.y = a.y-b.y;
}
};

int chacheng(vec val1,vec val2)
{

if( (val1.nag.x * val2.pos.y-val1.nag.y*val2.pos.x)> 0)
{

return 1;
}
else
return 0;
}

int main()
{
int num;
int a,b;

while(cin >> num && num!=0)
{
point *po = new point[num];
vec *ve = new vec[num];
for(int i = 0 ; i < num ; i++)
{
cin >> a >> b;
po[i].change(a,b);
}
for(int i = 0 ; i < num ; i++)
{
if(i != num-1)
ve[i].change(po[i],po[i+1]);
else
ve[i].change(po[i],po[0]);
}
int cnt = 0;
for(int i = 0 ; i < num ; i++)
{

if(i == num-1)
{
if(chacheng(ve[i],ve[0])==1)
{
cnt++;
}
}
else
{
if(chacheng(ve[i],ve[i+1])==1)
{
cnt++;
}
}
}
if(cnt == 0 || cnt == num)
{
cout << "convex" << endl;
}
else
{
cout << "concave"<<endl;
}
}
return 0;
}






代码改变世界