Box UVa 1587

盒子

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit  Status

Description

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 (  1w, h10 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

题意：给出六个矩形的长和宽，判断它们能否构成长方体的六个面。

代码：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define maxn 20
using namespace std;

struct Node
{
    int x,y;
}st[maxn];

int cmp(Node a,Node b)
{
    if (a.x==b.x)
        return a.yb) swap(a,b);
        st[0].x=a,st[0].y=b;
        for (int i=1;i<6;i++)
        {
            scanf("%d%d",&a,&b);
            if (a>b) swap(a,b);     //把较小的边放在前面
            st[i].x=a; st[i].y=b;
        }
        sort(st,st+6,cmp);      //排序
        for (int i=0;i<6;i+=2)
        {
            if (!(st[i].x==st[i+1].x&&st[i].y==st[i+1].y)) //如果每两个相邻的面的x和y不相等一定不满足
            {
                flag=0;
                break;
            }
        }
        if (flag)
        {
            if (st[0].x!=st[2].x)   //不同面的最小的两条边不能重合的不满足
                flag=0;
            if (!((st[0].y==st[4].x&&st[2].y==st[4].y)||(st[0].y==st[4].y&&st[2].y==st[4].x)))//判断第三个面能否与前两个面相接
                flag=0;
        }
        if (flag)
            printf("POSSIBLE\n");
        else
            printf("IMPOSSIBLE\n");
    }
    return 0;
}