poj 2007 Scrambled Polygon（凸多边形顶点输出）

题目：​​http://poj.org/problem?id=2007​​

描述：从(0，0)点开始输入一个凸多边形，这个凸多边形，占有三个象限，按照逆时针的方式输出各定点。

 

 

 

输出例子：

Sample Input

0 0
70 -50
60 30
-30 -50
80 20
50 -60
90 -20
-30 -40
-10 -60
90 10

Sample Output

(0,0)
(-30,-40)
(-30,-50)
(-10,-60)
(50,-60)
(70,-50)
(90,-20)
(90,10)
(80,20)
(60,30)

 

思路：从上图可以看出各象限都是斜率递增方式，建立4个vector对应四个象限，然后分别将各象限的点存储到相应的vector，最后对vector中的数据排序输出。（思路比较水）

 

代码：

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

struct Vertice{
  int x,y;
};

bool operator<(const Vertice &a, const Vertice &b)
{    
  return atan2((double)a.y, (double)a.x)<atan2((double)b.y, (double)b.x);
}

int main()
{
  Vertice point;
  vector<Vertice> ivec[4];
  int i=0;
  while(scanf("%d%d", &point.x, &point.y)!= EOF){
    //while(i++<10){
    //scanf("%d%d", &point.x, &point.y);
    if(point.x>0&&point.y>0)
      ivec[0].push_back(point);
    else if(point.x<0&&point.y>0)
      ivec[1].push_back(point);
    else if(point.x<0&&point.y<0)
      ivec[2].push_back(point);
    else if(point.x>0&&point.y<0)
      ivec[3].push_back(point);
  }
  for(int i=0;i<4;i++){
    if(ivec[i].size()!=0){
      sort(ivec[i].begin(),ivec[i].end());
    }
  }

  cout<<"(0,0)"<<endl;
  int begin;
  for(int i=0;i<4;i++){//找出凸边形的起始点
    if(ivec[i].size()==0){
      begin = (i+1)%4;
      break;
    }

  }

  for(int i=0;i<3;i++){
    int t = (begin+i)%4;
    if(ivec[t].size()!=0){
      vector<Vertice>::iterator it;
      for(it=ivec[t].begin();it!=ivec[t].end();it++){
        cout<<"("<<it->x<<","<<it->y<<")"<<endl;
      }
    }
  }

  system("pause");
  return 0;
}

 

​

View Code 

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

#define maxn 55
#define pi acos(-1)

struct Point
{
    int x, y;
} point[maxn];

bool operator <(const Point &a, const Point &b)
{
    return atan2(a.y, a.x) < atan2(b.y, b.x);
}

double cal(double a)
{
    if (a < 0)
        return a + 2 * pi;
    return a;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    scanf("%d%d", &point[0].x, &point[0].y);
    int n = 0;
    while (scanf("%d%d", &point[n].x, &point[n].y) != EOF)
        n++;
    sort(point, point + n);//按照artan角度排序
    double temp = 0;
    point[n] = point[0];
    int s;
    for (int i = 0; i < n; i++)
    {
        double a = cal(atan2(point[i + 1].y, point[i + 1].x) - atan2(point[i].y, point[i].x));//若前后两点之间的角度相差最大，这里就是凸边形的起始位置
        if (a > temp)
        {
            temp = a;
            s = (i + 1) % n;
        }
    }
    printf("(0,0)\n");
    for (int i = 0; i < n; i++)
        printf("(%d,%d)\n", point[(s + i) % n].x, point[(s + i) % n].y);
    return 0;
}