Box UVA - 1587

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

Sample Output

POSSIBLE

IMPOSSIBLE

HINT

题目的意思是给我们六个矩形的长和宽，让我们判断能否构成一个长方体。考虑长方形的特点可知，总会有两个面的长和宽hi相等的，同时12个边中总会又出现至少每4个边是相等的。可以根据这两个条件来判断输入样例。

Accepted

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

int cmp(const void* a, const void* b)

{

	return *(int*)a - *(int*)b;

}

int main()

{

	int a, b;

	while (scanf("%d %d", &a, &b) != EOF)

	{

		int arr[12] = { 0 };

		int arr2[6][2] = { 0 };

		arr2[0][0] = a > b ? a : b;

		arr2[0][1] = a > b ? b : a;

		int j = 2;

		arr[0] = a;arr[1] = b;

		for (int i = 1;i < 6;i++)

		{

			scanf("%d %d", &a, &b);

			arr2[i][0] = a > b ? a : b;

			arr2[i][1] = a > b ? b : a;

			arr[j++] = a;arr[j++] = b;

		}

		qsort(arr, 12, sizeof(int), cmp);

		int flag = 0;

		for (int i = 0;i < 12;i += 4)

			for (int j = 0;j < 4;j++)

				if (arr[i] != arr[j + i])

					flag = 1;

		if (!flag)

		{

			for (int i = 0;i < 6;i++)

			{

				int j;

				if (!arr2[i][0])continue;

				for (j = 0;j < 6;j++)

					if (i != j && arr2[i][0] == arr2[j][0] && arr2[i][1] == arr2[j][1])

					{

						arr2[i][0] = arr2[j][0] = arr2[i][1] = arr2[j][1] = 0;

						break;

					}

				if (j == 6)

				{

					flag = 1;

					break;

				}

			}

		}

		printf("%s\n", flag == 0 ? "POSSIBLE" : "IMPOSSIBLE");

	}

}

巴特西