Box
Description
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
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
在输入矩形的长和宽时,进行比较,大者赋值给前一个变量;
之后对数据排序;
分别判断1、2,3、4,5、6的矩形的长和宽能否构成一组矩形的面;
判断这三组平面能否构成一个矩形。
#include<stdio.h> struct pallet { int long1; int short2; }test[6]; int main() { int width, height; while(scanf("%d %d", &width, &height)!=EOF) { if(width<height) { int tmp = height; height = width; width = tmp; } test[0].long1 = width; test[0].short2 = height; int i; for(i=1; i<6; i++) { scanf("%d %d", &width, &height); if(width<height) { int tmp = height; height = width; width = tmp; } test[i].long1 = width; test[i].short2 = height; } for(i=0; i<6; i++) for(int j=5; j>i; j--) { if(test[j].long1<test[j-1].long1) { pallet cmp = test[j]; test[j] = test[j-1];; test[j-1] = cmp; } if((test[j].long1==test[j-1].long1)&&(test[j].short2<test[j-1].short2)) { pallet temp = test[j]; test[j] = test[j-1];; test[j-1] = temp; } } bool flag=true; for(i=0; i<3; i++) if((test[2*i].long1!=test[2*i+1].long1)||(test[2*i].short2!=test[2*i+1].short2)) { flag = false; break; } if(flag) if((test[3].long1!=test[5].long1)||(test[1].short2!=test[3].short2)||(test[1].long1!=test[5].short2)) flag = false; if(flag) printf("POSSIBLE\n"); else printf("IMPOSSIBLE\n"); } return 0; }