团体程序设计天梯赛(CCCC) L3012 水果忍者 上凸或下凹的证明

团体程序设计天梯赛代码。体现代码技巧，比赛技巧。  https://github.com/congmingyige/cccc_code

 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 #define ll long long
 const int maxn=1e4+;
 const int inf=1e9;
 const double eps=1e-;

 struct node
 {
     int x,y;
 }d1[maxn],d2[maxn],a[maxn];

 ll cross(node c,node a,node b)
 {
     ///两者相乘，也许大于int范围
     return 1ll*(c.y-a.y)*(c.x-b.x) - 1ll*(c.x-a.x)*(c.y-b.y);
 }

 bool cmp1(node a,node b)
 {
     ll v=cross(d1[],a,b);
     if (v==)
         return d1[].x-a.x<d1[].x-b.x;
     return v<;
 }

 bool cmp2(node a,node b)
 {
     ll v=cross(d2[],a,b);
     if (v==)
         return d2[].x-a.x<d2[].x-b.x;
     return v>;
 }

 /**
 两者在v==0处无法统一
 **/

 int main()
 {
     int n,x,ymax,ymin,i,j,ind=,m;
     double k,b;
     d1[].x=inf;
     scanf("%d",&n);
     for (i=;i<=n;i++)
     {
         scanf("%d%d%d",&x,&ymax,&ymin);
         d1[i]={x,ymax},d2[i]={x,ymin};
         if (d1[i].x<d1[ind].x)
             ind=i;
     }

     if (n==)
     {
         printf("%d %d %d %d",-,ymax,,ymax);
         return ;
     }

     swap(d1[],d1[ind]);
     sort(d1+,d1+n+,cmp1); ///注意是+2(对d1[2]~d1[n]排序)
     m=;
     for (i=;i<=n;i++)
     {
         while (m>= && cross(a[m],a[m-],d1[i])<=)
             m--;
         a[++m]=d1[i];
     }

     for (i=;i<=m;i++)
     {
         if (a[i].x<a[i-].x)
             break;
         k=1.0*(a[i].y-a[i-].y)/(a[i].x-a[i-].x);
         b=a[i].y-k*a[i].x;
         for (j=;j<=n;j++)
             if (k*d2[j].x+b<d2[j].y-eps)
                 break;
         if (j==n+)
         {
             printf("%d %d %d %d",a[i].x,a[i].y,a[i-].x,a[i-].y);
             return ;
         }
     }

     swap(d2[],d2[ind]);
     sort(d2+,d2+n+,cmp2);
     m=;
     for (i=;i<=n;i++)
     {
         while (m>= && cross(a[m],a[m-],d2[i])>=)
             m--;
         a[++m]=d2[i];
     }
     for (i=;i<=m;i++)
     {
         if (a[i].x<a[i-].x)
             break;
         k=1.0*(a[i].y-a[i-].y)/(a[i].x-a[i-].x);
         b=a[i].y-k*a[i].x;
         for (j=;j<=n;j++)
             if (k*d1[j].x+b>d1[j].y+eps)
                 break;
         if (j==n+)
         {
             printf("%d %d %d %d",a[i].x,a[i].y,a[i-].x,a[i-].y);
             return ;
         }
     }

     return ;
 }
 /*
 4
 -1 5 3
 2 4 3
 1 5 2
 0 4 3
 */