Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 876 Accepted Submission(s): 313
Special Judge
Problem Description
There are n boxes C1, C2, ..., Cn in 3Dspace. The edges of the boxes are parallel to the x, y or z-axis. We providesome relations of the boxes, and your task is to construct a set of boxessatisfying 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 Ci and Cjis positive.
X i j: The intersection volume is zero, andany point inside Ci has smaller x-coordinate than any point inside Cj.
Y i j: The intersection volume is zero, andany point inside Ci has smaller y-coordinate than any point inside Cj.
Z i j: The intersection volume is zero, andany point inside Ci has smaller z-coordinate than any point inside Cj.
.
Input
There will be at most 30 test cases. Eachcase 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 formatabove. The last test case is followed by n=R=0, which should not be processed.
Output
For each test case, print the case numberand either the word POSSIBLE or IMPOSSIBLE. If it's possible to construct theset 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. Theabsolute values of x1, y1, z1, x2, y2, z2 should not exceed 1,000,000.
Print a blank line after the output of eachtest case.
Sample Input
3 2
I 1 2
X 2 3
3 3
Z 1 2
Z 2 3
Z 3 1
1 0
0 0
Sample Output
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
Source
2009 Asia Wuhan Regional Contest Hosted byWuhan University
題解：
這道題是一道很好的題，在3D的空間中，長方體i在每一維看成是兩個點，如在x軸上，長方體i的兩個點是i和i+n（i表示左邊的點，i+n表示右邊的點，還未知道在x軸上的位置）。這道題給出了偏序關係,這個關係是小於關係。
1.I，兩個長方體x、y相交，在每個軸上都有x < y+n, y < x+n;
2.X，兩個長方體在X軸上，長方體x的兩個頂點都小於長方體的最左的頂點，即x+n < y；
3.Y，在y軸上有x+n<y;
4.Z，在Z軸上有x+n <y;
根據現有的偏序關係，用inx[]，iny[]，inz[]表示在特定軸上某個點的入度，即在現有條件下，某點比多少點大。而用vector vx[],vy[],vz[],記錄在特定軸上某點的出度的值，即在現有條件下比哪些點小，根據上面的偏序條件記錄後，進行拓撲排序，最後能將某軸上1到2*n個頂點拓撲完的，說明在該軸上，滿足條件。
源代碼：
#include <iostream>
#include <vector>
#include <stdio.h>
#include <string>
#include <queue>
#define MAX 2050
using namespace std;
int inx[MAX],iny[MAX],inz[MAX];
vector<int> vx[MAX],vy[MAX],vz[MAX];
int ansx[MAX],ansy[MAX],ansz[MAX];
int n,m;
void init()
{
memset(inx,0,sizeof(inx));
memset(iny,0,sizeof(iny));
memset(inz,0,sizeof(inz));
for(int i = 1;i <= 2*n;i++)
{
vx[i].clear();
vy[i].clear();
vz[i].clear();
}
for(int i = 1;i <= n;i++)
{
vx[i].push_back(i+n);
inx[i+n]++;
vy[i].push_back(i+n);
iny[i+n]++;
vz[i].push_back(i+n);
inz[i+n]++;
}
}
bool topologicalsort(int*in,vector<int> v[],int *ans)
{
intnum = 0;
queue<int>q;
for(int i = 1;i <= 2*n;i++)
{
if(!in[i])
q.push(i);
}
while(!q.empty())
{
intx = q.front();
q.pop();
ans[x] = num++;
for(int i = 0;i < v[x].size();i++)
{
intf = v[x][i];
in[f]--;
if(!in[f])
q.push(f);
}
}
if(num== 2*n)
returntrue;
else
returnfalse;
}
int main()
{
intcount = 0;
charc[5];
intx,y;
while(scanf("%d%d",&n,&m) != EOF &&n + m)
{
init();
for(int k = 0;k < m;k++)
{
scanf("%s%d%d",c,&x,&y);
if(c[0]== 'I')
{
vx[x].push_back(y+n);
inx[y+n]++;
vx[y].push_back(x+n);
inx[x+n]++;
vy[x].push_back(y+n);
iny[y+n]++;
vy[y].push_back(x+n);
iny[x+n]++;
vz[x].push_back(y+n);
inz[y+n]++;
vz[y].push_back(x+n);
inz[x+n]++;
}
elseif(c[0] == 'X')
{
vx[x+n].push_back(y);
inx[y]++;
}
elseif(c[0] == 'Y')
{
vy[x+n].push_back(y);
iny[y]++;
}
elseif(c[0] == 'Z')
{
vz[x+n].push_back(y);
inz[y]++;
}
}
intflag = 1;
if(!topologicalsort(inx,vx,ansx)){flag = 0;}
if(!topologicalsort(iny,vy,ansy)){flag = 0;}
if(!topologicalsort(inz,vz,ansz)){flag = 0;}
count++;
if(flag)
{
cout << "Case " << count <<": POSSIBLE" <<endl;
for(int i = 1;i <= n;i++)
cout<< ansx[i]<<" " <<ansy[i]<< " " <<ansz[i] << " "
<< ansx[i+n]<<" "<<ansy[i+n]<<" " << ansz[i+n] <<endl;
cout << endl;
}
else
{
cout << "Case "<<count <<": IMPOSSIBLE"<<endl<<endl;
}
}
return0;
}