凸包問題——圈水池(nyist 78)

來源:互聯網
上載者:User
#include <iostream>#include <cstring>#include <algorithm>using namespace std;struct point{int x,y;};int cmp1 (point A,point B)// 按y排序 {   return A.y < B.y || (A.y == B.y && A.x < B.x);  } int cmp2(point A,point B)//// 按x排序  {return A.x<B.x||(A.x==B.x&&A.y<B.y);  }int multi(point p0,point p1,point p2){return (p1.x-p0.x)*(p2.y-p0.y)>(p1.y-p0.y)*(p2.x-p0.x);  }//此處是大於符號 int graham(point pnt[], int n, point res[])//選中的點在儲存在res數組中,個數是top {   int i, len, k = 0, top = 1;     sort(pnt, pnt + n,cmp1);     if (n == 0) return 0; res[0] = pnt[0];     if (n == 1) return 1; res[1] = pnt[1];     if (n == 2) return 2; res[2] = pnt[2];     for (i = 2; i < n; i++)     {  while (top && multi(pnt[i], res[top], res[top-1])) top--;        res[++top] = pnt[i];     }     len = top; res[++top] = pnt[n - 2];     for (i = n - 3; i >= 0; i--)    {    while (top!=len && multi(pnt[i], res[top], res[top-1])) top--;          res[++top] = pnt[i];     }     return top;       // 返回凸包中點的個數 }int main(void){int t,i,n;point a[105],b[105];cin>>t;while(t--){   i=0;cin>>n;while(i<n) {   cin>>a[i].x>>a[i].y;   i++;  }n=graham(a,n,b);sort(b,b+n,cmp2);for(i=0;i<n;i++)cout<<b[i].x<<' '<<b[i].y<<endl;}}

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