「CF464B」Restore Cube

2014年9月8日3,2860

Peter had a cube with non-zero length of a side. He put the cube into three-dimensional space in such a way that its vertices lay at integer points (it is possible that the cube’s sides are not parallel to the coordinate axes). Then he took a piece of paper and wrote down eight lines, each containing three integers — coordinates of cube’s vertex (a single line contains coordinates of a single vertex, each vertex is written exactly once), put the paper on the table and left. While Peter was away, his little brother Nick decided to play with the numbers on the paper. In one operation Nick could swap some numbers inside a single line (Nick didn’t swap numbers from distinct lines). Nick could have performed any number of such operations.

When Peter returned and found out about Nick’s mischief, he started recollecting the original coordinates. Help Peter restore the original position of the points or else state that this is impossible and the numbers were initially recorded incorrectly.

Input

Each of the eight lines contains three space-separated integers — the numbers written on the piece of paper after Nick’s mischief. All numbers do not exceed 10⁶ in their absolute value.

Output

If there is a way to restore the cube, then print in the first line “YES“. In each of the next eight lines print three integers — the restored coordinates of the points. The numbers in the i-th output line must be a permutation of the numbers in i-th input line. The numbers should represent the vertices of a cube with non-zero length of a side. If there are multiple possible ways, print any of them.

If there is no valid way, print “NO” (without the quotes) in the first line. Do not print anything else.

Sample test(s)

input

0 0 0

0 0 1

0 1 1

1 1 1

output

YES

0 0 0

0 0 1

0 1 0

1 0 0

0 1 1

1 0 1

1 1 0

1 1 1

input

0 0 0

1 1 1

output

NO

题解

题意给定三维空间内8个点，每个点的xyz坐标被打乱，问是否能通过调整每个点的xyz坐标使得这八个点构成一个正方体。。。

我会告诉你我不知道正方体可以不平行于坐标轴wa飞了？

显然这题是个6^8搜索。。。

判断正方体可以通过任意一个顶点到其余七个顶点的距离判断

注意坐标全为0的情况

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#define pa pair<int,int>
#define inf 1000000000
#define ll long long 
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
bool flag;
int a[15][5],b[15][5],ans[15][5];
int c[6][3]={1,2,3,1,3,2,2,1,3,2,3,1,3,1,2,3,2,1};
ll dis(int x,int y)
{
    return (ll)(b[x][1]-b[y][1])*(b[x][1]-b[y][1])+(ll)(b[x][2]-b[y][2])*(b[x][2]-b[y][2])+(ll)(b[x][3]-b[y][3])*(b[x][3]-b[y][3]);
}
bool jud()
{
    ll a[10];
    for(int x=1;x<=8;x++)
    {   
        for(int i=1;i<=8;i++)
            a[i-1]=dis(i,x);
        sort(a,a+8);
        ll t=a[1];if(t==0)return 0;
        for(int i=1;i<=3;i++)if(a[i]!=t)return 0;
        for(int i=4;i<=6;i++)if(a[i]!=2*t)return 0;
        if(a[7]!=3*t)return 0;
    }
    return 1;
}
void dfs(int x)
{
    if(flag)return;
    if(x==9)
    {
        if(jud())
        {
            flag=1;
            for(int i=1;i<=8;i++)
                for(int j=1;j<=3;j++)
                    ans[i][j]=b[i][j];
        }
        return;
    }
    for(int i=0;i<6;i++)
    {
        for(int j=0;j<3;j++)
            b[x][j+1]=a[x][c[i][j]];
        dfs(x+1);
    }
}
int main()
{
    for(int i=1;i<=8;i++)
        for(int j=1;j<=3;j++)
            a[i][j]=read();
    dfs(1);
    if(flag)
    {
        puts("YES");
        for(int i=1;i<=8;i++)
        {
            for(int j=1;j<=3;j++)
                printf("%d ",ans[i][j]);
            puts("");
        }
    }
    else puts("NO");
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<map>

#include<set>

#include<vector>

#include<queue>

#define pa pair<int,int>

#define inf 1000000000

#define ll long long

using namespace std;

inline int read()

{

int x=0,f=1;char ch=getchar();

while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}

while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}

return x*f;

}

bool flag;

int a[15][5],b[15][5],ans[15][5];

int c[6][3]={1,2,3,1,3,2,2,1,3,2,3,1,3,1,2,3,2,1};

ll dis(int x,int y)

{

return (ll)(b[x][1]-b[y][1])*(b[x][1]-b[y][1])+(ll)(b[x][2]-b[y][2])*(b[x][2]-b[y][2])+(ll)(b[x][3]-b[y][3])*(b[x][3]-b[y][3]);

}

bool jud()

{

ll a[10];

for(int x=1;x<=8;x++)

{

for(int i=1;i<=8;i++)

a[i-1]=dis(i,x);

sort(a,a+8);

ll t=a[1];if(t==0)return 0;

for(int i=1;i<=3;i++)if(a[i]!=t)return 0;

for(int i=4;i<=6;i++)if(a[i]!=2*t)return 0;

if(a[7]!=3*t)return 0;

}

return 1;

}

void dfs(int x)

{

if(flag)return;

if(x==9)

{

if(jud())

{

flag=1;

for(int i=1;i<=8;i++)

for(int j=1;j<=3;j++)

ans[i][j]=b[i][j];

}

return;

}

for(int i=0;i<6;i++)

{

for(int j=0;j<3;j++)

b[x][j+1]=a[x][c[i][j]];

dfs(x+1);

}

int main()

{

for(int i=1;i<=8;i++)

for(int j=1;j<=3;j++)

a[i][j]=read();

dfs(1);

if(flag)

{

puts("YES");

for(int i=1;i<=8;i++)

{

for(int j=1;j<=3;j++)

printf("%d ",ans[i][j]);

puts("");

}

else puts("NO");

return 0;

}

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