HDU3231 Box Relations（拓撲排序）經典
來源：程序員人生 發布時間：2015-06-16 08:51:31 閱讀次數：4065次
Box Relations
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1042 Accepted Submission(s): 389
Special Judge
Problem Description
There are
n boxes
C1, C2, ..., Cn 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 Ci and Cj is positive.
- X i j: The intersection volume is zero, and any point inside Ci has smaller
x-coordinate than any point inside Cj.
- Y i j: The intersection volume is zero, and any point inside Ci has smaller
y-coordinate than any point inside Cj.
- Z i j: The intersection volume is zero, and any point inside Ci has smaller
z-coordinate than any point inside Cj.
.
Input
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.
Output
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.
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 by Wuhan University
題意：
在3維空間內，有n個長方體，棱都與坐標軸平行。給出1些關系，問是不是可能，若可能，輸出其中的1種。
關系有兩種：
1、I:兩個長方體有相交的體積。
2、X或Y或Z：某個長方體的所有點的某1維（X或Y或Z）的坐標完全小于另外一個長方體的任意1點。
#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++)//坐標X,Y,Z
for(int j=1;j<=n;j++)//盒子j,用兩個點表示，點j與點j+n
{
in[i][j+n]++; mapt[i][j].push_back(j+n); //點j的坐標比相應的點j+n坐標值小
}
}
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
");
else
{
printf("POSSIBLE
");
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("
");
}
}
printf("
");
}
}
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