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)。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <vector> 5 #include <cstring> 6 #include <string> 7 #include <algorithm> 8 #include <string> 9 #include <set> 10 #include <functional> 11 #include <numeric> 12 #include <sstream> 13 #include <stack> 14 #include <map> 15 #include <queue> 16 #pragma comment(linker, "/STACK:102400000,102400000") 17 #define CL(arr, val) memset(arr, val, sizeof(arr)) 18 19 #define ll long long 20 #define inf 0x7f7f7f7f 21 #define lc l,m,rt<<1 22 #define rc m + 1,r,rt<<1|1 23 #define pi acos(-1.0) 24 25 #define L(x) (x) << 1 26 #define R(x) (x) << 1 | 1 27 #define MID(l, r) (l + r) >> 1 28 #define Min(x, y) (x) < (y) ? (x) : (y) 29 #define Max(x, y) (x) < (y) ? (y) : (x) 30 #define E(x) (1 << (x)) 31 #define iabs(x) (x) < 0 ? -(x) : (x) 32 #define OUT(x) printf("%I64d\n", x) 33 #define lowbit(x) (x)&(-x) 34 #define Read() freopen("a.txt", "r", stdin) 35 #define Write() freopen("b.txt", "w", stdout); 36 #define maxn 1010 37 #define maxv 3000 38 #define mod 1000000000 39 using namespace std; 40 41 struct point 42 { 43 int x,y,id; 44 bool operator < (const point a) const 45 { 46 return x<a.x; 47 } 48 }p[10001]; 49 int main() 50 { 51 // freopen("a.txt","r",stdin); 52 int n,j,k; 53 scanf("%d",&n); 54 for(int i=1;i<=n;i++) 55 { 56 scanf("%d%d",&p[i].x,&p[i].y); 57 p[i].id=i; 58 } 59 sort(p+1,p+n+1); 60 double ans=0; 61 for(int i=2;i<=n;i++) 62 { 63 if(p[i].y==p[i-1].y) continue; 64 if((p[i].y-p[i-1].y)*1.0/(p[i].x-p[i-1].x)>ans) 65 { 66 ans=(p[i].y-p[i-1].y)*1.0/(p[i].x-p[i-1].x); 67 j=p[i].id; 68 k=p[i-1].id; 69 } 70 } 71 printf("%d %d\n",k,j); 72 return 0; 73 }
