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;}elsebreak;}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;}}