UVA - 1587 Box

最新推荐文章于 2021-03-20 18:26:36 发布

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本文介绍了一个简单的编程问题解决方案，通过输入六个长宽不等的托盘尺寸数据，判断是否能用这六个托盘组成一个箱子。文章提供了一段C++代码实现，包括数据输入、排序及比较逻辑。

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 inmillimeters 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.

SampleInput

1345 2584
2584 683
2584 1345
683 1345
683 1345
2584 683
1234 4567
1234 4567
4567 4321
4322 4567
4321 1234
4321 1234

SampleOutput

POSSIBLE

IMPOSSIBLE

分析：水题，不用多想，读取数据，排序，比较数据即可。

#include <stdio.h>
#include <algorithm>

using namespace std;

struct box{
	int x,y;
};
box b[6];

bool cmp(box a,box b)
{
	if(a.x==b.x)return a.y<b.y;
	return a.x<b.x;
}

int main()
{
	while(~scanf("%d",&b[0].x))
	{
		scanf("%d",&b[0].y);
		if(b[0].x>b[0].y)swap(b[0].x,b[0].y);
		for(int i=1; i<6; i++)
		{
			scanf("%d%d",&b[i].x,&b[i].y);
			if(b[i].x>b[i].y)swap(b[i].x,b[i].y);
		}

		sort(b,b+6,cmp);

		int error = 0;
		for(int i=0; i<3; i++)
		{
			if(b[i*2].x!=b[i*2+1].x || b[i*2].y!=b[i*2+1].y)
			{
				error = 1;
				break;
			}
		}
		if(!error && b[0].x==b[2].x && b[0].y==b[4].x && b[2].y==b[4].y)
			printf("POSSIBLE\n");
		else
			printf("IMPOSSIBLE\n");
	}
    return 0;
}