HDU3231 topology Ordering

Box relations

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 1334 Accepted Submission (s): 540
Special Judge

Problem Descriptionthere is NBoxes C1, C2, ..., CnIn 3D space. The edges of the boxes is parallel to the x, yOr Z-axis. We provide some relations of the boxes, and your task is to construct a set of boxes satisfying all these relations.

There is four kinds of relations (1 <= i,j<= N, Iis 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 have smaller x-coordinate than any PO int inside Cj.

• Y i j:the intersection volume is zero, and any point inside Ci have smaller y-coordinate than any PO int inside Cj.

• Z I j:the intersection volume is zero, and any point inside Ci have smaller z-coordinate than any point Inside Cj.

.

Inputthere'll is at the most test cases. Each case begins with a line containing the integers N(1 <= N<=) and R(0 <= R<= 100,000), the number of boxes and the number of relations. Each of the following RLines describes a relation, written in the format above. The last test case was followed by N= R=0, which should not be processed.

Outputfor 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 NLines contains six integers x1, y1, Z1, x2, Y2, Z2 ,That means the I-th box is the set of points ( x, y) satisfying x1 <= x <= x2, y1 <= y <= y2, z1 <= z <= Z2. The absolute values of x1, y1, Z1, x2, y2, Z2should not exceed 1,000,000.

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

Sample Input3 2I 1 2X 2 3Z 1 2Z 2 3Z 3 11 00 0

Sample outputcase 1:POSSIBLE0 0 0 2 2 1 1 3 3 8 8 9 9 9Case 2:impossiblecase 3:POSSIBLE0 0 0 1 1 1

Source2009 Asia Wuhan Regional Contest Hosted by Wuhan University test instructions: gives a positional relationship between n cubes, I, A, A and b for intersect, X, A, or X for a is less than the X coordinate of B, output Coordinate range code for the cube that meets the criteria:

//consider the six facets of a cube as 6 points, each with its own constraint (X1<X2,Y1<Y2,Z1<Z2), and the cube//constraints, so that three-dimensional coordinates are divided into three parts of the map, topological sorting, ranked in the back of more than 1 units in front of the length of the line.#include <iostream>#include<cstdio>#include<cstring>#include<queue>#include<vector>using namespacestd;Const intmaxn=2005;intNminch[3][maxn],val[3][maxn];vector<int>g[3][MAXN];voidinit () { for(intI=0;i<3; i++){         for(intj=1; j<=n*2; j + +){            inchI [j]=val[i][j]=0;        G[i][j].clear (); }    }     for(intI=0;i<3; i++){         for(intj=1; j<=n;j++){            inchI [j+n]++; G[i][j].push_back (J+N); }    }}BOOLtopo () { for(intI=0;i<3; i++) {Queue<int>Q; intCnt=0;  for(intj=1; j<=n;j++)        if(inchI [j]==0) {Val[i][j]=cnt++;        Q.push (j); }         while(!Q.empty ()) {            intA=Q.front (); Q.pop ();  for(intj=0;j< (int) G[i][a].size (); j + +){                intb=G[i][a][j]; VAL[I][B]=max (val[i][b],val[i][a]+1);//!                if(--inchI [b]==0) {Q.push (b); CNT++; }            }        }        if(cnt!=n*2)return 0; }    return 1;}intMain () {intcas=0;  while(SCANF ("%d%d", &n,&m) && (n+m))        {init (); Charch[5];intb;  while(m--) {scanf ("%s%d%d",ch,&a,&b); if(ch[0]=='I'){                 for(intI=0;i<3; i++){                    inchI [A+n]++;g[i][b].push_back (A +N); inchI [B+n]++;g[i][a].push_back (b +N); }            }            Else if(ch[0]=='X'){                inch[0][b]++;g[0][a+N].push_back (b); }            Else if(ch[0]=='Y'){                inch[1][b]++;g[1][a+N].push_back (b); }            Else{                inch[2][b]++;g[2][a+N].push_back (b); }} printf ("Case %d:",++CAs); if(Topo ()) {printf ("possible\n");  for(intI=1; i<=n;i++) printf ("%d %d%d%d%d%d\n", val[0][i],val[1][i],val[2][i],val[0][i+n],val[1][i+n],val[2][i+n]); }        Elseprintf"impossible\n"); printf ("\ n"); }    return 0;}

HDU3231 topology Ordering