uvalive 4728 Squares

题意：求所有正方形中两点距离最大值的平方值。

思路：旋转卡壳法。

分别用数组和vector存凸包时，旋转卡壳代码有所不同。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<memory.h>
#include<cstdlib>
#include<vector>
#define clc(a,b) memset(a,b,sizeof(a))
#define LL long long int
#define up(i,x,y) for(i=x;i<=y;i++)
#define w(a) while(a)
using namespace std;
const double inf=0x3f3f3f3f;
const int N = 4010;
const double eps = 5*1e-13;
const double PI = acos(-1.0);
using namespace std;

struct Point
{
    int x, y;
    Point(int x=0, int y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

int Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

int Dot(const Vector& A, const Vector& B)
{
    return A.x*B.x + A.y*B.y;
}

int Dist2(const Point& A, const Point& B)
{
    return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
}

bool operator < (const Point& p1, const Point& p2)
{
    return p1.x < p2.x || (p1.x == p2.x && p1.y < p2.y);
}

bool operator == (const Point& p1, const Point& p2)
{
    return p1.x == p2.x && p1.y == p2.y;
}

// 点集凸包
// 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
// 注意：输入点集会被修改
vector<Point> ConvexHull(vector<Point>& p)
{
    // 预处理，删除重复点
    sort(p.begin(), p.end());
    p.erase(unique(p.begin(), p.end()), p.end());

    int n = p.size();
    int m = 0;
    vector<Point> ch(n+1);
    for(int i = 0; i < n; i++)
    {
        while(m > 1 && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    for(int i = n-2; i >= 0; i--)
    {
        while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;
        ch[m++] = p[i];
    }
    if(n > 1) m--;
    ch.resize(m);
    return ch;
}

// 返回点集直径的平方
int diameter2(vector<Point>& points)
{
    vector<Point> p = ConvexHull(points);
    int n = p.size();
    if(n == 1) return 0;
    if(n == 2) return Dist2(p[0], p[1]);
    p.push_back(p[0]); // 免得取模
    int ans = 0;
    for(int u = 0, v = 1; u < n; u++)
    {
        // 一条直线贴住边p[u]-p[u+1]
        for(;;)
        {
            // 当Area(p[u], p[u+1], p[v+1]) <= Area(p[u], p[u+1], p[v])时停止旋转
            // 即Cross(p[u+1]-p[u], p[v+1]-p[u]) - Cross(p[u+1]-p[u], p[v]-p[u]) <= 0
            // 根据Cross(A,B) - Cross(A,C) = Cross(A,B-C)
            // 化简得Cross(p[u+1]-p[u], p[v+1]-p[v]) <= 0
            int diff = Cross(p[u+1]-p[u], p[v+1]-p[v]);
            if(diff <= 0)
            {
                ans = max(ans, Dist2(p[u], p[v])); // u和v是对踵点
                if(diff == 0) ans = max(ans, Dist2(p[u], p[v+1])); // diff == 0时u和v+1也是对踵点
                break;
            }
            v = (v + 1) % n;
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        vector<Point> points;
        for(int i = 0; i < n; i++)
        {
            int x, y, w;
            scanf("%d%d%d", &x, &y, &w);
            points.push_back(Point(x, y));
            points.push_back(Point(x+w, y));
            points.push_back(Point(x, y+w));
            points.push_back(Point(x+w, y+w));
        }
        printf("%d
", diameter2(points));
    }
    return 0;
}