n个点中求任意两点组成斜率的最大值

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1100

首先按x坐标排序,然后相邻的三个点A,B,C 组成的三条直线必然有K(AC)<max(K(A,B),K(B,C));(K是斜率)

所以斜率一定会在相邻的两点中产生,排序O(nlogn),更新最大值O(n)。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1010
 #define maxv 3000
 #define mod 1000000000
 using namespace std;

 struct point
 {
     int x,y,id;
     bool operator < (const point a) const
     {
         return x<a.x;
     }
 }p[];
 int main()
 {
     // freopen("a.txt","r",stdin);
     int n,j,k;
     scanf("%d",&n);
     for(int i=;i<=n;i++)
     {
         scanf("%d%d",&p[i].x,&p[i].y);
         p[i].id=i;
     }
     sort(p+,p+n+);
     double ans=;
     for(int i=;i<=n;i++)
     {
         if(p[i].y==p[i-].y) continue;
         if((p[i].y-p[i-].y)*1.0/(p[i].x-p[i-].x)>ans)
         {
             ans=(p[i].y-p[i-].y)*1.0/(p[i].x-p[i-].x);
             j=p[i].id;
             k=p[i-].id;
         }
     }
     printf("%d %d\n",k,j);
     return ;
 }