poj2187 Beauty Contest 凸包 Graham扫描

Beauty Contest

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 35619 Accepted: 11040

Description

Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 ... 10,000. No two farms share the same pair of coordinates. 

Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm 

Output

* Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other. 

Sample Input

40 00 11 11 0

Sample Output

2

Hint

Farm 1 (0, 0) and farm 3 (1, 1) have the longest distance (square root of 2) 

题目大意：

平面上有n个牧场，i号牧场的位置在格点（xi,yi），所有牧场的位置互不相同，请计算距离最远的两个牧场的距离，输出最远距离的平方。

思路：

显然暴力枚举每两个点的距离会超时，需要避免计算一些不必要的点对。

由于最远点对一定在凸包上，所以只要枚举凸包上所有点对，计算其距离即可。

凸包算法有很多，这里用比较易于实现的Graham扫描算法。

#include <cstdio>  #include <string> #include <iostream>#include <cstring>#include <cctype>#include <algorithm>#include <functional>#include <cstdlib>#include <list>#include <utility>#include <vector>#include <bitset>using namespace std;typedef long long LL;const double EPS = 1e-6;//考虑误差的加法运算double Add(double a, double b) {    if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;    return a + b;}//二维向量结构体struct Point2 {    double x, y;    Point2() {}    Point2(double x, double y) : x(x), y(y) {}    Point2 operator+(const Point2& p) const {        return Point2(Add(x, p.x), Add(y, p.y));    }    Point2 operator-(const Point2& p) const {        return Point2(Add(x, -p.x), Add(y, -p.y));    }    Point2 operator*(double d) {        return Point2(x*d, y*d);    }    //內积    double Dot(const Point2& p) const {        return Add(x*p.x, y*p.y);    }    //外积    double Det(const Point2& p) const {        return Add(x*p.y, -y*p.x);    }};bool CmpXY(const Point2& p, const Point2& q) {    if (p.x != q.x) return p.x < q.x;    return p.y < q.y;}//求凸包vector<Point2> GrahamScan(Point2* ps, int n) {    sort(ps, ps + n, CmpXY);    int k = 0; //凸包的顶点数    vector<Point2> qs(n << 1); //构造中的凸包    //构造凸包的下侧    for (int i = 0; i < n; ++i) {        while (k > 1 && (qs[k - 1] - qs[k - 2]).Det(ps[i] - qs[k - 1]) <= 0) --k;        qs[k++] = ps[i];    }    //构造凸包上侧    for (int i = n - 2, t = k; i >= 0; --i) {        while (k > t && (qs[k - 1] - qs[k - 2]).Det(ps[i] - qs[k - 1]) <= 0) --k;        qs[k++] = ps[i];    }    qs.resize(k - 1);    return qs;}double Dist(const Point2& p, const Point2& q) {    return (p - q).Dot(p - q);}const int maxn = 5e4 + 5;int n;Point2 ps[maxn];int main() {#ifdef NIGHT_13    freopen("in.txt", "r", stdin);#endif    std::ios::sync_with_stdio(false);    while (scanf("%d", &n) == 1) {        for (int i = 0; i < n; ++i) {            scanf("%lf%lf", &ps[i].x, &ps[i].y);        }        vector<Point2> ConvexHull = GrahamScan(ps, n);        double ans = 0;        for (int i = 0; i < ConvexHull.size(); ++i) {            for (int j = 0; j < i; ++j) {                ans = max(ans, Dist(ConvexHull[i], ConvexHull[j]));            }        }        printf("%.0f\n", ans);    }    return 0;}