凸包问题

凸包问题

 

     凸包基础知识见：http://baike.baidu.com/view/707209.htm 或者，

参考《国际大学生程序设计竞赛例题解 》（一）计算几何部分和LRJ的黑书。

研究了两天的凸包问题，也在POJ上找了几道题目练习了一下，这里给出自己写的

Graham扫描法求凸包的模板\源代码：/*--------------------------------------------------------------------// Abstract: Graham扫描法求凸包                                        // Author:王莉峰(Fandywang)                                            // History:2007-11-21                                                  // Version:Microsoft Visual C++6.0                                     //--------------------------------------------------------------------*/#include #include #include typedef struct CPoint{ double x, y;}Point;const int MAXSIZE = 10000;const double EPS = 1E-6;int  stack[MAXSIZE]; //数组模拟堆栈，存放凸包的逆序顶点int  top;            //凸包顶点数p[stack[0]...stack[top]]Point p[MAXSIZE];void GrahamScan(const int &n); //Graham扫描法求凸包double Dist(const Point &arg1, const Point &arg2); //两点距离double Length(const int &n);   //多边形周长double Area(const int &n);     //多边形面积void Swap(const int &i, const int &j); //交换两个点double Multi(const Point &p0, const Point &p1, const Point &p2); //叉积运算int cmp(const void *arg1, const void *arg2); //自定义排序:按照极角int main(void){ int  n;// freopen("Input.txt", "r", stdin); while( scanf("%d", &n), n ) {  int  i;  for( i=0; i < n; ++i )  {   scanf("%lf %lf", &p[i].x, &p[i].y);  }  if( n == 1 )  {   printf("周长: 0.00\t面积: 0.00\n");  }  else if( n == 2 )   {   printf("周长: %.2lf\t面积: 0.00\n", Dist(p[0], p[1]));  }  else   {    GrahamScan(n);    /* for( int k=0; k <= top; ++k )    {     printf("%lf  %lf\n", p[stack[k]].x, p[stack[k]].y);    }*/    printf("周长: %.2lf\t面积: %.2lf\n", Length(top+1), Area(top+1));  } } return 0;}double Length(const int &n){ double  sum = Dist( p[stack[n-1]], p[stack[0]] ); for( int i = 0; i < n-1; ++i ) {  sum += Dist( p[stack[i]], p[stack[i+1]] ); } return sum;}double Area(const int &n){ double area = 0; for( int i=0; i < n-1; ++i ) {  area += (p[stack[i]].x*p[stack[i+1]].y) - (p[stack[i+1]].x*p[stack[i]].y); } area = fabs(area) / 2; return area;}double Dist(const Point &arg1, const Point &arg2){ return sqrt( (arg1.x - arg2.x)*(arg1.x - arg2.x) + (arg1.y - arg2.y)*(arg1.y - arg2.y) );}void Swap(const int &i, const int &j){ Point temp = p[i]; p[i] = p[j]; p[j] = temp;}//通过矢量叉积求极角关系(p0p1)(p0p2)//P*Q > 0,P在Q的顺时针方向... ...double Multi(const Point &p0, const Point &p1, const Point &p2){ return (p1.x-p0.x)*(p2.y-p0.y) - (p2.x-p0.x)*(p1.y-p0.y);}int cmp(const void *arg1, const void *arg2){ Point a = *(Point*)arg1; Point b = *(Point*)arg2; double k = Multi(p[0], a, b); if( k < -EPS ) return 1; else if( fabs(k) < EPS && (Dist(a, p[0]) - Dist(b, p[0])) > EPS ) return 1; else return -1;}void GrahamScan(const int &n){ int  i, u = 0; for( i = 1; i < n; ++i ) {//找到最左下的顶点, 令其为p[0]  if( ( p[i].y < p[u].y ) || ( p[i].y == p[u].y && p[i].x < p[u].x ) ) u = i; } i = 0; Swap(i, u); //只交换一次 //将点按照相对p[0]的极角从小到大排序, 用qsort qsort( p+1, n-1, sizeof(p[0]), cmp ); //将每一个点逐一检验, 是凸包点放入栈中, 不是则弹出栈。 //注意: 排序后, p[0]和p[1]一定是凸包的顶点 stack[0] = 0; stack[1] = 1; stack[2] = 2;    top = 2; for( int j = 3; j < n; ++j ) {  while( Multi( p[j], p[stack[top]], p[stack[top-1]] ) > 0 )  {   --top;   if( top == 0 ) break;  }  stack[++top] = j; }}

//凸包题目：
POJ 1113 Wall  凸包的基本应用
POJ 1228 Grandpa's Estate 凸包拓展
POJ 2187 Beauty Contest 用凸包求最远点对
提高题目：
POJ 1794 Castle Walls 
POJ 2007 Scrambled Polygon