ZOJ-2099 矩形边界

2099:给出一段折线，求一个能包住次折线的最小矩形。

思路：找出最左最右的x值,最上最下的y值即可。
由于整数范围内都合法，初始时有些麻烦。因此设了一个变量代表第一次赋值。

#include<stdio.h>
#include<iostream>
using namespace std;


int main()
{		
	int southwest_x;
	int southwest_y;
	int northeast_x;
	int northeast_y;
	int x;
	int y;
	bool isend=false;
	bool isfirst;

	while(!isend)
	{
		southwest_x=0;
		southwest_y=0;
		northeast_x=0;
		northeast_y=0;
		isfirst=true;
	
		while(1)
		{
			cin>>x;
			cin>>y;
			if(x==0&&y==0)
			{
				if(!(southwest_x||southwest_y||northeast_x||northeast_y))
				{
					isend=true;
					break;
				}
				else
					break;
			}

			if(isfirst)
			{
				southwest_x = x;
				southwest_y = y; 
				northeast_x = x;
				northeast_y = y;
				isfirst=false;
			}

			if(x<=southwest_x)
				southwest_x = x;

			if(y<=southwest_y)
				southwest_y = y; 

			if(x>=northeast_x)
				northeast_x=x;

			if(y>=northeast_y)
				northeast_y=y;
			
		}
		if(!isend)
			cout<<southwest_x<<" "<<southwest_y<<" "<<northeast_x<<" "<<northeast_y<<endl;
	}

}