紫书第三章训练 UVA 1587 Box by 16 BobHuang

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. 
Input 
Input 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 ≤ 10 000) — width and height of the pallet in 
millimeters respectively. 
Output 
For 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 Input 
1345 2584 
2584 683 
2584 1345 
683 1345 
683 1345 
2584 683 
1234 4567 
1234 4567 
4567 4321 
4322 4567 
4321 1234 
4321 1234

Sample Output 
POSSIBLE 
IMPOSSIBLE

这道题就是检验啊，检验给你的6个面是不是长方体，这6个面你要单独保存会比较好？我是没有，但是我感觉那样思路应该更清晰，其实也就是两两配对，拿flag标记下，
然后就wa了，想了想长方体三边是a,b,c那么每两组之间必有两个数会想等的，加上这个条件就AC了。

#include <bits/stdc++.h>
using namespace std;
int main()
{int a[12];
while(cin>>a[0]>>a[1]>>a[2]>>a[3]>>a[4]>>a[5]>>a[6]>>a[7]>>a[8]>>a[9]>>a[10]>>a[11]){
    sort(a,a+2);
    sort(a+2,a+4);
    sort(a+4,a+6);
    sort(a+6,a+8);
    sort(a+8,a+10);
    sort(a+10,a+12);
    int f[6];
    for(int i=0;i<6;i++)
    f[i]=1;
    int s=1;
    for(int i=0;i<12;i+=2){
        if(f[i/2])
        for(int j=i+2;j<12;j+=2)
        if(f[j/2]){
            if(a[i]==a[j]&&a[i+1]==a[j+1])
                f[i/2]=f[j/2]=0;
            else if(!(a[i]==a[j]||a[i+1]==a[j+1]||a[i]==a[j+1]||a[i+1]==a[j]))
                s=0;
        }
    }
    for(int i=0;i<6;i++)
    if(f[i]) {s=0;break;}
    if(s)printf("POSSIBLE
");
    else printf("IMPOSSIBLE
");

}
return 0;}