POJ 2977 Box walking

Description

You are given a three-dimensional box of integral dimensions l_x × l_y × l_z The edges of the box are axis-aligned, and one corner of the box is located at position (0, 0, 0). Given the coordinates (x, y, z) of some arbitrary position on the surface of the box, your goal is to return the square of the length of the shortest path along the box’s surface from (0, 0, 0) to (x, y, z).

If l_x = 1, l_y = 2, and l_z = 1, then the shortest path from (0, 0, 0) to (1, 2, 1) is found by walking from (0, 0, 0) to (1, 1, 0), followed by walking from (1, 1, 0) to (1, 2, 1). The total path length is √8.

Input

The input test file will contain multiple test cases, each of which consists of six integers l_x, l_y, l_z, x, y, z where 1 ≤ l_x, l_y, l_z ≤ 1000. Note that the box may have zero volume, but the point (x, y, z) is always guaranteed to be on one of the six sides of the box. The end-of-file is marked by a test case with l_x = l_y = l_z = x = y = z = 0 and should not be processed.

Output

For each test case, write a single line with a positive integer indicating the square of the shortest path length. (Note: The square of the path length is always an integer; thus, your answer is exact, not approximate.)

Sample Input

1 1 2 1 1 2
1 1 1 1 1 1
0 0 0 0 0 0

Sample Output

8
5

#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
#include<queue>
#include<stack>
#include<algorithm>
#include<vector>
#include<cmath>
#include<string>
#include<functional>
using namespace std;
const int maxn=505;
int lx,ly,lz,x,y,z,ans;

int f(int x) {return x*x;}

int main()
{
	while (cin>>lx>>ly>>lz>>x>>y>>z,x+y+z+lx+ly+lz)
	{
		ans=0x7FFFFFFF;
		if (x==0||y==0||z==0) ans=f(x)+f(y)+f(z);
		else 
		if (x==lx) 
		{
			ans=min(ans,min(f(x+y)+f(z),f(x+z)+f(y)));
			ans=min(ans,min(f(y+lz)+f(lx+lz-z),f(z+ly)+f(lx+ly-y)));
		}
		else 
		if (y==ly) 
		{
			ans=min(ans,min(f(y+x)+f(z),f(y+z)+f(x)));
			ans=min(ans,min(f(x+lz)+f(ly+lz-z),f(z+lx)+f(ly+lx-x)));
		}
		else
		if (z==lz)
		{
			ans=min(ans,min(f(z+y)+f(x),f(z+x)+f(y)));
			ans=min(ans,min(f(y+lx)+f(lx+lz-x),f(x+ly)+f(ly+lz-y)));
		}
		printf("%d\n",ans);
	}
	return 0;
}