uva 1587 box

Ivan works at a factory that produces heavy machinery. He has a simple job — he knocks up wooden boxes of different sizes to pack machinery for delivery to the customers. Each box is a rectangular parallelepiped. Ivan uses six rectangular wooden pallets to make a box. Each pallet is used for one side of the box.Joe delivers pallets for Ivan. Joe is not very smart and often makes mistakes — he brings Ivan pallets that do not fit together to make a box. But Joe does not trust Ivan. It always takes a lot of time to explain Joe that he has made a mistake. Fortunately, Joe adores everything related to computers and sincerely believes that computers never make mistakes. Ivan has decided to use this for his own advantage. Ivan asks you to write a program that given sizes of six rectangular pallets tells whether it is possible to make a box out of them.InputInput file contains several test cases. Each of them consists of six lines. Each line describes one pallet and contains two integer numbers w and h (1 ≤ w,h ≤ 10000) — width and height of the pallet in millimeters respectively.OutputFor each test case, print one output line. Write a single word ‘POSSIBLE’ to the output file if it is possible to make a box using six given pallets for its sides. Write a single word ‘IMPOSSIBLE’ if it is not possible to do so.Sample Input1345 2584 2584 683 2584 1345 683 1345 683 1345 2584 683 1234 4567 1234 4567 4567 4321 4322 4567 4321 1234 4321 1234Sample OutputPOSSIBLE IMPOSSIBLE

这道题得要注意是长方体 要3条相邻的边不一样 才是正确的   这里首先就行排序 让后再比较 

<span style="font-size:24px;">下面是源码  </span>

<span style="font-size:24px;">#include<stdio.h>int main(){ int temp,i=0,j,a[6],b[6],flag; while(scanf("%d%d",&a[0],&b[0])!=EOF) {  if(a[0]<b[0]){temp=a[0];a[0]=b[0];b[0]=temp;}  for(i=1;i<6;i++)  {   scanf("%d%d",&a[i],&b[i]);   if(a[i]<b[i]){temp=a[i];a[i]=b[i];b[i]=temp;}  }  flag=0;  for(i=0;i<5;i++)   for(j=i+1;j<6;j++)   {    if((a[i]<a[j]) || (a[i]==a[j] && b[i]<b[j]))    {     temp=a[i];a[i]=a[j];a[j]=temp;     temp=b[i];b[i]=b[j];b[j]=temp;    }   }  for(i=0;i<6;i+=2)   if(a[i]!=a[i+1]||b[i]!=b[i+1])flag=1;        if(a[0]!=a[2]||b[0]!=a[4]||b[2]!=b[4])flag=1;     if(flag)printf("IMPOSSIBLE\n");  else printf("POSSIBLE\n"); } return 0;}</span>