题目链接:http://poj.org/problem?id=2187
第一个最远点对。求的是最远点对的平方。
求最远点对,我们应该知道一条性质,最远点对,一定是这一系列点的凸包上的两个点。这道题,直接枚举凸包上的点就可以。当然,旋转卡壳也是可以的。
直接求凸包+枚举点。OK。
Code:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> using namespace std; const int N = 5 * 1e4 + 5; const double eps = 1e-8; struct POINT{ double x, y; }p[N], st[N]; int n, top; double cross(POINT o, POINT a, POINT b){ return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x); } double Distance(POINT a, POINT b){ return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y); } bool cmp(POINT a, POINT b){ if(a.y < b.y) return true; else if(a.y == b.y && a.x < b.x) return true; return false; } bool cmp1(POINT a, POINT b){ if(cross(p[0], a, b) > eps) return true; else if(fabs(cross(p[0], a, b)) < eps && fabs(Distance(p[0], a) - Distance(p[0], b)) > eps)return true; return false; } bool Judge(POINT a, POINT b, POINT c){ if(cross(a, b, c) > eps) return true; else if(fabs(cross(a, b, c)) < eps && Distance(b, a) < Distance(b, c)) return true; else return false; } void Graham_scan(){ sort(p, p + n, cmp); sort(p + 1, p + n, cmp1); top = 0; st[top ++] = p[0]; st[top ++] = p[1]; for(int i = 2; i < n; i ++){ while(top >= 2 && Judge(st[top - 1], st[top - 2], p[i])) top --; st[top ++] = p[i]; } } double ALL(){ double ans = 0; for(int i = 0; i < top - 1; i ++){ for(int j = i + 1; j < top; j ++) if(Distance(st[i], st[j]) - ans > eps) ans = Distance(st[i], st[j]); } return ans; } int main(){ // freopen("1.txt", "r", stdin); while(~scanf("%d", &n)){ for(int i = 0; i < n; i ++){ scanf("%lf %lf", &p[i].x, &p[i].y); } Graham_scan(); printf("%.0lf\n", ALL()); } return 0; }
旋转卡壳还是需要学习的,枚举点在某些问题上的时间复杂度太高。