UVA - 1587 Box 麻烦

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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.

$\epsfbox{p3214.eps}$

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 $\le$ w, h $\le$ 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 25842584 6832584 1345683 1345683 13452584 6831234 45671234 45674567 43214322 45674321 12344321 1234

Sample Output

POSSIBLEIMPOSSIBLE#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>int main(){    double a[3],b[3],c[3],x,y;    int i,sign;    while(~scanf("%lf%lf",&x,&y))    {        sign=0;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        a[1]=x,a[2]=y,a[0]=1;        for(i=1;i<6;i++)        {            scanf("%lf%lf",&x,&y);            if((x==a[1]&&y==a[2])||(x==a[2]&&y==a[1]))            a[0]++;            else            {                if(b[1]<=1e-6&&b[2]<=1e-6)                b[1]=x,b[2]=y,b[0]=1;                else                {                    if((x==b[1]&&y==b[2])||(x==b[2]&&y==b[1]))                    b[0]++;                    else                    {                        if(c[1]<=1e-6&&c[2]<=1e-6)                        c[1]=x,c[2]=y,c[0]=1;                        else                        {                            if((x==c[1]&&y==c[2])||(x==c[2]&&y==c[1]))                            c[0]++;                            else                            {                                sign=1;                            }                        }                    }                }            }        }        if(sign)        printf("IMPOSSIBLE\n");        else        {            if(a[0]==6&&fabs(a[1]-a[2])<1e-6)            printf("POSSIBLE\n");            else if(b[0]==4&&fabs(a[1]-a[2])<1e-6||a[0]==4&&fabs(b[1]-b[2])<1e-6)            printf("POSSIBLE\n");            else if(a[0]==2&&b[0]==2&&c[0]==2)            {                if(fabs(a[1]-b[1])<1e-6||fabs(a[2]-b[2])<1e-6||fabs(a[2]-b[1])<1e-6||fabs(a[1]-b[2])<1e-6)                {                    if(fabs(a[1]-b[1])<1e-6)                    {                        if((fabs(a[2]-c[2])<1e-6&&fabs(c[1]-b[2])<1e-6)||(fabs(a[2]-c[1])<1e-6&&fabs(c[2]-b[2])<1e-6))                        printf("POSSIBLE\n");                        else                        printf("IMPOSSIBLE\n");                    }                    else if(fabs(a[1]-b[2])<1e-6)                    {                        if((fabs(a[2]-c[2])<1e-6&&fabs(c[1]-b[1])<1e-6)||(fabs(a[2]-c[1])<1e-6&&fabs(c[2]-b[1])<1e-6))                        printf("POSSIBLE\n");                        else                        printf("IMPOSSIBLE\n");                    }                    else if(fabs(a[2]-b[1])<1e-6)                    {                        if((fabs(a[1]-c[2])<1e-6&&fabs(b[2]-c[1])<1e-6)||(fabs(a[1]-c[1]<1e-6&&fabs(b[2]-c[2])<1e-6)))                        printf("POSSIBLE\n");                        else                        printf("IMPOSSIBLE\n");                    }                    else if(fabs(a[2]-b[2])<1e-6)                    {                        if((fabs(a[1]-c[2])<1e-6&&fabs(b[1]-c[1])<1e-6)||(fabs(a[1]-c[1]<1e-6&&fabs(b[1]-c[2])<1e-6)))                        printf("POSSIBLE\n");                        else                        printf("IMPOSSIBLE\n");                    }                }                else                printf("IMPOSSIBLE\n");            }            else            printf("IMPOSSIBLE\n");        }    }    return 0;}

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