UVA-1587 Box

最新推荐文章于 2021-03-20 18:26:36 发布

Yvetteweiss

最新推荐文章于 2021-03-20 18:26:36 发布

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文章标签： UVA

本文链接：https://blog.youkuaiyun.com/yvetteweiss/article/details/60324722

本文介绍了一个程序设计问题，即通过给定六个矩形托盘的尺寸来判断是否可以组装成一个长方体包装箱。文章详细展示了一段C语言代码实现，包括结构体定义、排序和比较算法等步骤。

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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.
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 ≤ 10000) — 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 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

分析：本体重点在于如何判断两个矩形全等，以及是否能组成长方体（不同矩形的长与宽必有一组相等）。

Source:

#include<stdio.h>
#include<stdlib.h>
struct pallet
{
    int w,h;
};
int is_same(struct pallet a,struct pallet b)
{
    if(a.w==b.w&&a.h==b.h)
        return 1;
    else
        return 0;
}
int main()
{
    struct pallet p[6],temp;
    int i,j,t;
    while(scanf("%d%d",&p[0].w,&p[0].h)!=EOF)
    {
        if(p[0].w<p[0].h)
        {
            t=p[0].w;
            p[0].w=p[0].h;
            p[0].h=t;
        }
        for(i=1;i<6;i++)
        {
            scanf("%d%d",&p[i].w,&p[i].h);
            if(p[i].w<p[i].h)
            {
                t=p[i].w;
                p[i].w=p[i].h;
                p[i].h=t;
            }
        }
        for(i=0;i<5;i++)
            for(j=0;j<5-i;j++)
                if(p[j].w<p[j+1].w)
                {
                    temp=p[j];
                    p[j]=p[j+1];
                    p[j+1]=temp;
                }
        for(i=0;i<5;i++)
            for(j=0;j<5-i;j++)
                if(p[j].w==p[j+1].w&&p[j].h<p[j+1].h)
                {
                    temp=p[j];
                    p[j]=p[j+1];
                    p[j+1]=temp;
                }
        for(i=0;i<6;i+=2)
            if(!is_same(p[i],p[i+1]))
                break;
        if(i==6&&(p[0].w==p[2].w)&&((p[4].w==p[0].h&&p[4].h==p[2].h)||(p[4].w==p[2].h&&p[4].h==p[0].h)))
            printf("POSSIBLE\n");
        else
            printf("IMPOSSIBLE\n");
    }
    return 0;
}