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 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

解题思路: 把w和h作比较，然后w是小的，h是大的，排好一级排序后，按照6个面来排序，排好之后如果是possible，就会有0==1的面，2==3的面，4==5的面，取出不同的3个面，然后进行宽高是否相等。

#include <iostream>
#include <cstdio>
#include <algorithm>
#define MAXD 10
using namespace std;

struct M{
    int w;
    int h;
}rectangular[MAXD];
//比较交换
bool cmp(M m,M n){
    if(m.w > n.w) return true;
    if(m.w < n.w) return false;
    if(m.h > n.h) return true;
    else return false;
}
//判断是否能构造长方体
bool solve(){
    M ans[MAXD];
    int m = 0;
    for(int i = 0; i < 6 ; i+=2){
        if(rectangular[i].w==rectangular[i+1].w && rectangular[i].h==rectangular[i+1].h)
            ans[m] = rectangular[i],m++;
        else return false;
    }
    if (rectangular[0].h==rectangular[2].h && rectangular[0].w==rectangular[4].h && rectangular[2].w==rectangular[4].w)
        return true;
    else return false;
}

void exchange(int i){ //宽高交换
    int temp = rectangular[i].w;
    rectangular[i].w = rectangular[i].h;
    rectangular[i].h = temp;
}

int main(){   //一级排序是宽高排序
    while(scanf("%d",&rectangular[0].w) != EOF){
        scanf("%d",&rectangular[0].h);
        if(rectangular[0].w > rectangular[0].h) exchange(0);
        for(int i = 1; i < 6 ; i++){
            scanf("%d%d",&rectangular[i].w,&rectangular[i].h);
            if(rectangular[i].w > rectangular[i].h) exchange(i);
        }
        sort(rectangular,rectangular+6,cmp);//二级排序把面排序
        bool ok = solve();
        ok? printf("POSSIBLE\n"):printf("IMPOSSIBLE\n");
    }
}