HDU3231 Box Relations（拓扑排序）经典
来源：程序员人生 发布时间：2015-06-16 08:51:31 阅读次数：3702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
题意：
在3维空间内，有n个长方体，棱都与坐标轴平行。给出1些关系，问是不是可能，若可能，输出其中的1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1维（X或Y或Z）的坐标完全小于另外一个长方体的任意1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
");
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("
");
}
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
