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.

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 25842584 6832584 1345683 1345683 13452584 6831234 45671234 45674567 43214322 45674321 12344321 1234

Sample Output 

POSSIBLEIMPOSSIBLE

比较水，能确定长方形的判定条件就可以了

AC代码：

#include<cstdio>#include<algorithm>using namespace std;const int N = 6;struct rec{ int l, w; } r[N];bool cmp(rec a, rec b){return a.w < b.w || (a.w == b.w && a.l < b.l);}int main(){int a, b, ok;while (~scanf("%d%d", &r[0].w, &r[0].l)){ok = 1;if (r[0].w > r[0].l) swap(r[0].w, r[0].l);for (int i = 1; i < 6; ++i){scanf("%d%d", &r[i].w, &r[i].l);if (r[i].w > r[i].l) swap(r[i].w, r[i].l);}sort(r, r + N, cmp);for (int i = 0; i < N; i += 2)if (r[i].w != r[i + 1].w || r[i].l != r[i + 1].l) ok = 0;if (r[0].w != r[2].w || r[0].l != r[4].w || r[2].l != r[4].l) ok = 0;puts(ok ? "POSSIBLE" : "IMPOSSIBLE");}return 0;}