nyoj-78-圈水池（Graham算法求凸包）

题目链接

 /*
     Name:nyoj-78-圈水池
     Copyright:
     Author:
     Date: 2018/4/27 9:52:48
     Description:
     Graham求凸包
     zyj大佬的模板，改个输出就能用
 */
 #include <cstring>
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 using namespace std;
 struct point{
     double x, y;
 };
 bool mult(point sp, point ep, point op){ //叉积
     return (sp.x - op.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - op.y);
 }
 inline bool operator < (const point &l, const point &r){
      return l.y < r.y || (l.y == r.y && l.x < r.x);
 }
 int graham(int n, point pnt[], point res[]){
     int i, len, top = ;
     sort(pnt, pnt + n);//排序找最小的点
     if (n == ){
         return ;
     }
     res[] = pnt[];
     if (n == ){
         return ;
     }
     res[] = pnt[];
     if (n == ){
         return ;
     }
     res[] = pnt[];
     for (i = ; i < n; i++){
         while (top && mult(pnt[i], res[top], res[top - ])){
             top--;
         }
         res[++top] = pnt[i];
     }
     len = top;
     res[++top] = pnt[n - ];
     for (i = n - ; i >= ; i--){
         while (top != len && mult(pnt[i], res[top], res[top - ])){
             top--;
         }
         res[++top] = pnt[i];
     }
     return top;  // 返回凸包中点的个数
 }
 bool cmp(const point &l, const point &r){
      return l.x < r.x || (l.x == r.x && l.y < r.y);
 }
 int main()
 {
     int t;
     cin>>t;
     point pnt[], res[];
     while (t--)  {
         memset(res, , sizeof(res));
         memset(pnt, , sizeof(pnt));
         int m;
         cin>>m;
         for (int i=; i<m; i++)
             cin>>pnt[i].x>>pnt[i].y;
         int ct = graham(m, pnt, res);
         sort(res, res + ct, cmp);
         for (int i=; i<ct; i++) {
             printf("%.0f %.0f\n", res[i].x, res[i].y);
         }
     }
     return ;
 }