poj 2007 Scrambled Polygon（凸多边形顶点输出）

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：

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#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;
}

#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;
}

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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", &

补充：软件开发 , C++ ,