UVA 11800 - Determine the Shape(计算几何)
来源：程序员人生 发布时间：2015-04-09 08:38:31 阅读次数：3662次
题意：给定4个点，判断形状
思路：先求个凸包，就可以把4个点排序，然后就是利用几何去判断，利用点积判垂直，利用叉积判平行
还有这题有个坑啊，明明说好是没有点共线的，实际上是有的，所以求凸包如果不是4个点，直接输出不规则4边形便可
代码：
#include <cstdio>
#include <cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN = 4;
struct Point {
int x, y;
Point() {}
Point(double x, double y) {
this->x = x;
this->y = y;
}
void read() {
scanf("%d%d", &x, &y);
}
} list[MAXN], p[MAXN];
typedef Point Vector;
Vector operator + (Vector A, Vector B) {
return Vector(A.x + B.x, A.y + B.y);
}
Vector operator - (Vector A, Vector B) {
return Vector(A.x - B.x, A.y - B.y);
}
Vector operator * (Vector A, double p) {
return Vector(A.x * p, A.y * p);
}
Vector operator / (Vector A, double p) {
return Vector(A.x / p, A.y / p);
}
int Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;} //点积
int Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;} //叉积
int stack[MAXN], top;
int xmult(Point p0, Point p1, Point p2) {
return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);
}
double dist(Point p1,Point p2) {
return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
}
int dist2(Point p1, Point p2) {
return (p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y);
}
//极角排序函数 ， 角度相同则距离小的在前面
bool cmp(Point p1,Point p2) {
int tmp= xmult(list[0], p1, p2);
if(tmp > 0) return true;
else if(tmp == 0 && dist(list[0], p1) < dist(list[0], p2)) return true;
else return false;
}
bool LineParallel(Vector v, Vector w) {
return Cross(v, w) == 0;
}
bool LineVertical(Vector v, Vector w) {
return Dot(v, w) == 0;
}
//输入并把最左下方的点放在list[0]并且进行极角排序
void init(int n) {
int i, k;
Point p0;
scanf("%d%d", &list[0].x, &list[0].y);
p0.x = list[0].x;
p0.y = list[0].y;
k = 0;
for(i = 1; i < n; i++) {
scanf("%d%d", &list[i].x, &list[i].y);
if((p0.y > list[i].y) || ((p0.y == list[i].y) && (p0.x > list[i].x))) {
p0.x = list[i].x;
p0.y = list[i].y;
k = i;
}
}
list[k] = list[0];
list[0] = p0;
sort(list + 1, list + n, cmp);
}
void graham(int n) {
int i;
if(n == 1) {top = 0; stack[0] = 0;}
if(n == 2) {
top = 1;
stack[0] = 0;
stack[1] = 1;
}
if(n > 2) {
for(i = 0; i <= 1; i++) stack[i] = i;
top = 1;
for(i = 2; i < n; i++) {
while(top > 0 && xmult(list[stack[top - 1]], list[stack[top]], list[i]) <= 0) top--;
top++;
stack[top]=i;
}
}
for (int i = 0; i <= top; i++)
p[i] = list[stack[i]];
}
int t;
void solve() {
if (LineParallel(p[1] - p[0], p[2] - p[3]) && LineParallel(p[2] - p[1], p[3] - p[0])) {
if (LineVertical(p[1] - p[0], p[2] - p[1]) && LineVertical(p[2] - p[1], p[3] - p[2]) && LineVertical(p[0] - p[3], p[3] - p[2])) {
if (dist2(p[1], p[0]) == dist2(p[2], p[1])) printf("Square
");
else printf("Rectangle
");
} else {
if (dist2(p[1], p[0]) == dist2(p[2], p[1])) printf("Rhombus
");
else printf("Parallelogram
");
}
} else if (LineParallel(p[1] - p[0], p[2] - p[3]) || LineParallel(p[2] - p[1], p[3] - p[0])) printf("Trapezium
");
else printf("Ordinary Quadrilateral
");
}
int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
init(4);
graham(4);
printf("Case %d: ", ++cas);
if (top < 3) {
printf("Ordinary Quadrilateral
");
continue;
}
for (int i = 0; i < 4; i++)
p[i] = list[stack[i]];
solve();
}
return 0;
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
