FZU 2148 ——Moon Game 判断凸四边形

题目链接:点击打开链接

题目大意：给你一堆的点，然后判断最多能组成多少个凸四边形

解题思路：因为最多有30个点，所以直接暴力搜索就可以了

所学知识点： 如何判断凸四边形，在这种有坐标的情况下当然是用对角线相交来做了

                        如何判断对角线相交呐：如果两条线段相交，则这两条线段必然相互相互跨立，所以用两次叉积都小于等于0来做就可以了

在判断是不是凸四边形的时候还要注意，只要有一对对角线相交就可以，所以还要枚举一下对角线

//能否组成四边形问题//判断对角线的香蕉情况，当然不是用包凸啊,xixixi//用叉积判断对角线的相交情况//如果相交则必然在两侧#include<iostream>#include<cstdio>#include<cstring>using namespace std;class Point{public:    int x, y;    Point():x(0), y(0) {}    Point(int xx, int yy): x(xx), y(yy) {}};Point p[35];int n;void init(){    cin >> n;    for(int i = 0; i < n; i++)        cin >> p[i].x >> p[i].y;}bool judge1(Point a, Point b, Point c, Point d) //判断两条线段是否相交{   long long m1 = (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);   long long m2 = (b.x - a.x) * (d.y - a.y) - (d.x - a.x) * (b.y - a.y);   long long m3 = (d.x - c.x) * (a.y - c.y) - (a.x - c.x) * (d.y - c.y);   long long m4= (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);    if(m1*m2 <= 0 && m3 *m4 <= 0) return true;    else return false;}bool judge2(int i, int j, int k) //判断3个点是否在一条直线上{    int x1 = p[i].x, x2 = p[j].x, x3 = p[k].x;    int y1 = p[i].y, y2 = p[j].y, y3 = p[k].y;    if((y2-y1)*(x3-x2) == (y3-y2)*(x2-x1)) return false;    else return true;}int solve(){    int cnt = 0;    for(int i = 0; i < n; i++)    {        for(int j = i+1; j < n; j++)        {            for(int k = j+1; k < n; k++)            {                for(int m = k + 1; m < n; m++)                {//                    if(judge(i, j, k) == false) continue;//                    if(judge(i, j, m) == false) continue;//                    if(judge(i, k, m) == false) continue;//                    if(judge(j, k, m) == false) continue;//                    cout << i << " " << j << " " << k  << " " << m << endl;                    if(judge1(p[i], p[j], p[k], p[m])||inter(p[i], p[m], p[k], p[j]) || inter(p[i], p[k], p[j], p[m]))                    cnt++;                }            }        }    }    return cnt;}int main(){    int T;    scanf("%d", &T);    for(int i = 1; i <= T; i++)    {        init();        int ans = solve();        cout << "Case " << i << ": " << ans << endl;    }    return 0;}