poj 1228 凸包

题目链接：http://poj.org/problem?id=1228

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

const double eps = 1e-8;
const double PI = acos(-1.0);
const double INF = 1000000000000000.000;

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 * (double p,Vector A){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);
}

inline 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);
inline double Dot(Vector A, Vector B){ return A.x*B.x + A.y*B.y; }
inline double Length(Vector A)    { return sqrt(Dot(A,A)); }
inline double Angle(Vector A, Vector B)  { return acos(Dot(A,B) / Length(A) / Length(B)); }
inline 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--;   //如果希望在凸包的边上有输入点，把<= 换成 <.
        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;
}

bool IsPointOnSegment(Point P,Point A,Point B){
    return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(P-A,P-B)) < 0;
}


/*************************************分 割 线*****************************************/

const int maxn = 1005;

Point P[maxn];
Point Goal[maxn];

int main()
{
    freopen("E:\acm\input.txt","r",stdin);

    int T;
    cin>>T;

    while(T--){
        int n;
        cin>>n;

        for(int i=0;i<n;i++){
            scanf("%lf %lf",&P[i].x,&P[i].y);
        }


        int cnt = ConvexHull(P,n,Goal);

        if(cnt <= 2 || cnt*2 > n || n < 6){
            printf("NO
");
            continue;
        } cout<<cnt<<"  "<<Goal[cnt].x<<"  "<<Goal[cnt].y<<endl;
        Goal[cnt] = Goal[0];
        int i;
        for(i=0;i<cnt;i++){
            bool flag = false;
            for(int j=0;j<n;j++){
               // if(j == i || j == i+1) continue;    加了这句就WA 7 次，痛苦啊
                if(IsPointOnSegment(P[j],Goal[i],Goal[i+1])){
                    flag = true;
                }
            }
            if(!flag) break;
        }
        if(i<cnt) printf("NO
");
        else      printf("YES
");
    }
}