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

0 070 -5060 30-30 -5080 2050 -6090 -20-30 -40-10 -6090 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;}

 

网友：http://www.cnblogs.com/rainydays/archive/2011/09/02/2163310.html

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;}