UVA 11800 - Determine the Shape(计算几何)

题意：给定四个点，判断形状

思路：先求个凸包，就能把四个点排序，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，其实是有的，所以求凸包如果不是4个点，直接输出不规则四边形即可

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\n");            else printf("Rectangle\n");        } else {            if (dist2(p[1], p[0]) == dist2(p[2], p[1])) printf("Rhombus\n");            else printf("Parallelogram\n");        }    } else if (LineParallel(p[1] - p[0], p[2] - p[3]) || LineParallel(p[2] - p[1], p[3] - p[0])) printf("Trapezium\n");    else printf("Ordinary Quadrilateral\n");}int main() {    int cas = 0;    scanf("%d", &t);    while (t--) {        init(4);        graham(4);        printf("Case %d: ", ++cas);        if (top < 3) {            printf("Ordinary Quadrilateral\n");            continue;        }        for (int i = 0; i < 4; i++)            p[i] = list[stack[i]];        solve();    }    return 0;}