There are n boxes C₁, C₂, ..., C_n in 3D space. The edges of the boxes are parallel to the x, y or z-axis. We provide some relations of the boxes, and your task is to construct a set of boxes satisfying all these relations.

There are four kinds of relations (1 <= i,j <= n, i is different from j):

• I i j: The intersection volume of C_i and C_j is positive.
  
• X i j: The intersection volume is zero, and any point inside C_i has smaller x-coordinate than any point inside C_j.
  
• Y i j: The intersection volume is zero, and any point inside C_i has smaller y-coordinate than any point inside C_j.
  
• Z i j: The intersection volume is zero, and any point inside C_i has smaller z-coordinate than any point inside C_j.

.

There will be at most 30 test cases. Each case begins with a line containing two integers n (1 <= n <= 1,000) and R (0 <= R <= 100,000), the number of boxes and the number of relations. Each of the following R lines describes a relation, written in the format above. The last test case is followed by n=R=0, which should not be processed.

For each test case, print the case number and either the word POSSIBLE or IMPOSSIBLE. If it‘s possible to construct the set of boxes, the i-th line of the following n lines contains six integers x1, y1, z1, x2, y2, z2, that means the i-th box is the set of points (x,y,z) satisfying x1 <= x <= x2, y1 <= y <= y2, z1 <= z <= z2. The absolute values of x1, y1, z1, x2, y2, z2 should not exceed 1,000,000.

Print a blank line after the output of each test case.

3 2
I 1 2
X 2 3
3 3
Z 1 2
Z 2 3
Z 3 1
1 0
0 0

Case 1: POSSIBLE
0 0 0 2 2 2
1 1 1 3 3 3
8 8 8 9 9 9

Case 2: IMPOSSIBLE

Case 3: POSSIBLE
0 0 0 1 1 1

题意：

在三维空间内，有n个长方体，棱都与坐标轴平行。给出一些关系，问是否可能，若可能，输出其中的一种。

关系有两种：

1、I:两个长方体有相交的体积。

2、X或Y或Z：某个长方体的所有点的某一维（X或Y或Z）的坐标完全小于另一个长方体的任意一点。

#include<stdio.h>
#include<vector>
#include<queue>
using namespace std;
const int N = 2005;

vector<int>mapt[3][N];
int in[3][N],v[3][N],n;

void init()
{
    for(int i=0;i<3;i++)
        for(int j=1;j<=n*2;j++)
        in[i][j]=v[i][j]=0,mapt[i][j].clear();

    for(int i=0;i<3;i++)//坐标X,Y,Z
        for(int j=1;j<=n;j++)//盒子j,用两个点表示，点j与点j+n
        {
            in[i][j+n]++; mapt[i][j].push_back(j+n); //点j的坐标比相应的点j+n坐标值小
        }
}

int topeSort()
{
    for(int i=0;i<3;i++)
    {
        int k=0;
        queue<int>q;
        for(int j=1;j<=n;j++)
            if(in[i][j]==0)
            q.push(j),v[i][j]=k++;

        while(!q.empty())
        {
            int s=q.front(); q.pop();
            int len=mapt[i][s].size();
            for(int j=0; j<len; j++)
            {
                int tj=mapt[i][s][j];
                in[i][tj]--;
                if(v[i][s]+1>v[i][tj])
                    v[i][tj]=v[i][s]+1;
                if(in[i][tj]==0)
                    q.push(tj),k++;
            }
        }
        if(k!=n*2)
            return 0;
    }
    return 1;
}
int main()
{
    char ch[5];
    int m,a,b,cas=0;
    while(scanf("%d%d",&n,&m)>0&&n+m!=0)
    {
        init();
        while(m--)
        {
            scanf("%s%d%d",ch,&a,&b);
            if(ch[0]=='I')
            {
                for(int i=0;i<3;i++)
                {
                    mapt[i][a].push_back(b+n); in[i][b+n]++;
                    mapt[i][b].push_back(a+n); in[i][a+n]++;
                }
            }
            else if(ch[0]=='X')
                mapt[0][a+n].push_back(b),in[0][b]++;
            else if(ch[0]=='Y')
                mapt[1][a+n].push_back(b),in[1][b]++;
            else if(ch[0]=='Z')
                mapt[2][a+n].push_back(b),in[2][b]++;
        }

        printf("Case %d: ",++cas);
        if(topeSort()==0)
            printf("IMPOSSIBLE\n");
        else
        {
            printf("POSSIBLE\n");
            for(int i=1;i<=n;i++)
            {
                printf("%d",v[0][i]);
                for(int j=1;j<3;j++)
                    printf(" %d",v[j][i]);
                for(int j=0;j<3;j++)
                    printf(" %d",v[j][i+n]);
                printf("\n");
            }
        }
        printf("\n");
    }
}