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 ;

 }