UVa1453或La4728 凸包+枚举（或旋转卡壳）

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4199

没想透为啥旋转卡壳跟枚举跑时间差不多。n太小吧！

枚举法：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 100050;
const int maxe = 100000;
const int INF = 0x3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
typedef Point Vector;

Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A , double p){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);}

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

int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b){
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

///向量（x,y）的极角用atan2(y,x);
double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
double Length(Vector A)    { return Dot(A,A); }
double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B)  { return A.x*B.y - A.y * B.x; }

//凸包：
/**Andrew算法思路：首先按照先x后y从小到大排序（这个地方没有采用极角逆序排序，所以要进行两次扫描），删除重复的点后得到的序列p1,p2.....,然后把p1和p2放到凸包中。从p3开始，当新的
点在凸包“前进”方向的左边时继续，否则依次删除最近加入凸包的点，直到新点在左边；**/

//Goal[]数组模拟栈的使用；
int ConvexHull(Point* P,int n,Point* Goal){
    sort(P,P+n);
    int m = unique(P,P+n) - P;    //对点进行去重；
    int cnt = 0;
    for(int i=0;i<m;i++){       //求下凸包；
        while(cnt>1 && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0)  cnt--;
        Goal[cnt++] = P[i];
    }
    int temp = cnt;
    for(int i=m-2;i>=0;i--){     //逆序求上凸包；
        while(cnt>temp && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0) cnt--;
        Goal[cnt++] = P[i];
    }
    if(cnt > 1) cnt--;  //减一为了去掉首尾重复的；
    return cnt;
}
/*********************************分割线******************************/

Point P[maxn*4],Goal[maxn*4];
int n;

int main()
{
    //freopen("E:\acm\input.txt","r",stdin);
    int T;
    cin>>T;
    while(T--){
        cin>>n;
        int cnt = 0;
        double x,y,w;
        for(int i=1;i<=n;i++){
            scanf("%lf %lf %lf",&x,&y,&w);
            P[cnt++] = Point(x,y);
            P[cnt++] = Point(x+w,y);
            P[cnt++] = Point(x,y+w);
            P[cnt++] = Point(x+w,y+w);
        }
        cnt = ConvexHull(P,cnt,Goal);
        double Maxlen = 0;
        for(int i=0;i<cnt;i++)
           for(int j=i+1;j<cnt;j++){
             Maxlen = max(Maxlen,Length(Goal[j]-Goal[i]));
        }
        printf("%.lf
",Maxlen);
    }
    return 0;
}

旋转卡壳：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 100500;
const int maxe = 100000;
const int INF = 0x3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);

struct Point{
    double x,y;
    Point(double x=0, double y=0) : x(x),y(y){ }    //构造函数
};
typedef Point Vector;

Vector operator + (Vector A , Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A , Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A , double p){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A , double p){return Vector(A.x/p,A.y/p);}

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

int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
bool operator == (const Point& a, const Point& b){
    return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

///向量（x,y）的极角用atan2(y,x);
double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
double Length(Vector A)    { return Dot(A,A); }
double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B)  { return A.x*B.y - A.y * B.x; }

//凸包：
/**Andrew算法思路：首先按照先x后y从小到大排序（这个地方没有采用极角逆序排序，所以要进行两次扫描），删除重复的点后得到的序列p1,p2.....,然后把p1和p2放到凸包中。从p3开始，当新的
点在凸包“前进”方向的左边时继续，否则依次删除最近加入凸包的点，直到新点在左边；**/
// 如果不希望在凸包的边上有输入点，把两个 <= 改成 <
//Goal[]数组模拟栈的使用；
int ConvexHull(Point* P,int n,Point* Goal){
    sort(P,P+n);
    int m = unique(P,P+n) - P;    //对点进行去重；
    int cnt = 0;
    for(int i=0;i<m;i++){       //求下凸包；
        while(cnt>1 && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0)  cnt--;
        Goal[cnt++] = P[i];
    }
    int temp = cnt;
    for(int i=m-2;i>=0;i--){     //逆序求上凸包；
        while(cnt>temp && dcmp(Cross(Goal[cnt-1]-Goal[cnt-2],P[i]-Goal[cnt-2])) <= 0) cnt--;
        Goal[cnt++] = P[i];
    }
    if(cnt > 1) cnt--;  //减一为了去掉首尾重复的；
    return cnt;
}
//旋转卡壳可以用于求凸包的直径、宽度，两个不相交凸包间的最大距离和最小距离
//计算凸包直径，输入凸包Goal，顶点个数为n，按逆时针排列，输出直径的平方
double RotatingCalipers(Point* Goal,int n){
    double ret = 0;
    Goal[n]=Goal[0];  //补上使凸包成环；
    int pv = 1;
    for(int i=0;i<n;i++){   //枚举边Goal[i]Goal[i+1],与最远顶点Goal[pv];利用叉积求面积的方法求最大直径；；
        while(fabs(Cross(Goal[i+1]-Goal[pv+1],Goal[i]-Goal[pv+1]))>fabs(Cross(Goal[i+1]-Goal[pv],Goal[i]-Goal[pv])))
              pv = (pv+1)%n;
        ret=max(ret,max(Length(Goal[i]-Goal[pv]),Length(Goal[i+1]-Goal[pv+1]))); //这个地方不太好理解，就是要考虑当pv与pv+1所在直线平行于i与i+1的情况；
    }
    return ret;
}
/*********************************分割线******************************/

Point P[maxn*4],Goal[maxn*4];
int n;

int main()
{
    //freopen("E:\acm\input.txt","r",stdin);
    int T;
    cin>>T;
    while(T--){
        cin>>n;
        int cnt = 0;
        double x,y,w;
        for(int i=1;i<=n;i++){
            scanf("%lf %lf %lf",&x,&y,&w);
            P[cnt++] = Point(x,y);
            P[cnt++] = Point(x+w,y);
            P[cnt++] = Point(x,y+w);
            P[cnt++] = Point(x+w,y+w);
        }
        cnt = ConvexHull(P,cnt,Goal);
        double Maxlen = RotatingCalipers(Goal,cnt);
        printf("%.lf
",Maxlen);
    }
    return 0;
}