poj 2007 Scrambled Polygon（凸包Graham扫描线模板）

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Scrambled Polygon

题目链接：http://poj.org/problem?id=2007

代码如下：

#include<stdio.h>#include<limits.h>#include<stdlib.h>#include<string.h>#include<deque>#include<algorithm>using namespace std;typedef struct{int x;int y;}Point;int n, temp;Point top1, top2, s[105];deque<Point> q;bool comp2(Point a, Point b)//按幅角排序{if((a.x-top1.x)*(b.y-top1.y)-(a.y-top1.y)*(b.x-top1.x)>0)return 1;else if((a.x-top1.x)*(b.y-top1.y)-(a.y-top1.y)*(b.x-top1.x)==0 && abs(a.x-top1.x)<abs(b.x-top1.x))return 1;return 0;}int main(void){n=1;int ax,ay;ax = ay = INT_MAX;while(~scanf("%d%d", &s[n].x, &s[n].y)){    n++;    if(s[n].y<ay || s[n].y==ay && s[n].x<ax)    //每次记录最左下角的，先下后左            temp = n, ay = s[n].y, ax = s[n].x;}    top1 = s[temp]; //作为sort排序的比较标准    sort(s+1, s+n, comp2);    q.push_back(s[1]);    q.push_back(s[2]);    temp = 3;    while(temp<n)    {        top1 = q.back();        q.pop_back();//临时弹出用作判定        top2 = q.back();        if((top2.x-top1.x)*(s[temp].y-top1.y)-(top2.y-top1.y)*(s[temp].x-top1.x)<=0)        {            q.push_back(top1);//符合条件,临时弹出的归队            q.push_back(s[temp]);//当前点入队            temp++;//下一个点作为当前点        }        //不符合条件丢弃    }    int flag=0;    while(q.empty()==0)    {        top1 = q.front();        if((top1.x==0 && top1.y==0) || flag)//根据题意，循环入栈出栈，判断从(0,0)开始输出        {            printf("(%d,%d)\n", top1.x, top1.y);            q.pop_front();            flag=1;        }        else if(!flag)        {            q.pop_front();            q.push_back(top1);        }    }return 0;}

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