UVA 1587 - Box

1587 - Box

Time limit: 3.000 seconds

Ivan works at a factory that produces heavy machinery. He has a simple job — he knocks up woodenboxes of different sizes to pack machinery for delivery to the customers. Each box is a rectangularparallelepiped. Ivan uses six rectangular wooden pallets to make a box. Each pallet is used for one sideof the box.

Joe delivers pallets for Ivan. Joe is not very smart and often makes mistakes — he brings Ivanpallets that do not fit together to make a box. But Joe does not trust Ivan. It always takes a lot oftime to explain Joe that he has made a mistake

Fortunately, Joe adores everything related to computers and sincerely believes that computers nevermake mistakes. Ivan has decided to use this for his own advantage. Ivan asks you to write a programthat 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 palletand contains two integer numbers w and h (1 ≤ w, h ≤ 10 000) — width and height of the pallet inmillimeters respectively.

Output

For each test case, print one output line. Write a single word ‘POSSIBLE’ to the output file if it ispossible to make a box using six given pallets for its sides. Write a single word ‘IMPOSSIBLE’ if it is notpossible 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

题目大意就是说，给出六个长方形的长宽，求这六个长方形能否组成一个长方体。六条边排序之后比较一下。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{    int x,y;};node tran[10];bool cmp(node a,node b){    if(a.x!=b.x)        return a.x>b.x;    else        return a.y>b.y;}void rswap(int &a,int &b){    a=a^b;    b=b^a;    a=a^b;}int main(){    while(~scanf("%d",&tran[0].x))    {        scanf("%d",&tran[0].y);        if(tran[0].x<tran[0].y)            rswap(tran[0].x,tran[0].y);        for(int i=1;i<6;++i)        {            scanf("%d%d",&tran[i].x,&tran[i].y);            if(tran[i].x<tran[i].y)                rswap(tran[i].x,tran[i].y);        }        sort(tran,tran+6,cmp);        int flag=1;        for(int i=0;i<3;++i)        {            if(tran[2*i].x==tran[2*i+1].x&&tran[2*i].y==tran[2*i+1].y)                flag=1;            else            {                flag=0;                break;            }        }        if(!flag)        {            printf("IMPOSSIBLE\n");            continue;        }        if(tran[0].x!=tran[2].x)            flag=0;        else if(tran[0].y!=tran[4].x&&tran[0].y!=tran[4].y)            flag=0;        else        {            if(tran[0].y==tran[4].x&&tran[2].y!=tran[4].y)                flag=0;            else if(tran[0].y==tran[4].y&&tran[2].y!=tran[4].x)                flag=0;        }        if(!flag)            printf("IMPOSSIBLE\n");        else            printf("POSSIBLE\n");    }    return 0;}