poj 1696 Space Ant

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题意：对顶点圈出一个向左转的最长的罗璇, 顶点只能走过一次，罗璇没有交线;

分析：对任意点集, 都能圈出一个罗璇包含所有节点; 用叉积判断罗璇方向,圈出罗璇就行,注意输出点数和起始位置的判定

#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <vector>#include <queue>#define MAX 10005#define INF 100000000#include <iostream>#define PI acos(-1.0)#define eps 1e-8#define INF 1e8using namespace std;struct Point{    int index;    int x, y;    Point() {}    Point(int xx, int yy):x(xx), y(yy){}    bool operator < (const Point & p)const {        if(x != p.x)            return ( x < p.x);        return y < p.y;    }    void input(){        cin>>index>>x>>y;    }    void output(){            cout<<x<<" "<<y<<endl;    }    int xmult(const Point &vec)const{        return x * vec.y - y * vec.x;    }    Point sub(const Point &p)const{        return Point(x - p.x, y - p.y);    }}p[MAX];vector<Point>ansArray;int M, N;void init(){        cin>>N;        for(int i = 0; i < N; i++){            p[i].input();        }        ansArray.clear();       sort(p, p + N);}void getAnsArray(){    bool v[MAX];    memset(v, 0, sizeof(v));    int m = 0, mMin = 1, j = 0, sign = 1, miny = INF;    for(int i = 0; i < N; i++){            if(p[i].y < miny){                miny = p[i].y;                j = i;            }    }    while(true){ //串完所有点        for( ; j < N && j >= 0; j+= sign){            if(!v[p[j].index]){                while(m > mMin && (ansArray[m -1].sub(ansArray[m - 2]).xmult(p[j].sub(ansArray[m - 2]))) < 0){                    m --;                    ansArray.pop_back();                }                m++;                ansArray.push_back(p[j]);            }        }        if(m >= N)break;        mMin = m;         sign *= -1;         j += sign;        for(int jj = 0; jj < ansArray.size(); jj++){            v[ansArray[jj].index] = true;        }    }    cout<<ansArray.size();    for(int i = 0; i < ansArray.size(); i++){        cout<<" "<<ansArray[i].index;    }cout<<endl;}int main(){    cin>>M;    while(M--){        init();        getAnsArray();    }    return 0;}