bzoj4466 Super Cube

Description

Hypercube is the extension of a cube in a high-dimensional space (it degrades to a square in 2-D, 1
degenerate into segments in the case of dimensions). In the field of theoretical computer science, hypercube can often be used with 2 binary
coded together. The will nature of research on theoretical computer Science also has
own thinking.

is the graph of the hypercube in the 0~4 dimension space. Obviously we can put the hypercube
each vertex of the body is regarded as a point, and each edge is viewed as a side, so that a graph is given, and I
They call it a hypercube chart.
the hypercube map in D-dimensional space has 2D dots, and we numbered the points from 0 to 2d-1.
There is an interesting and important conclusion: there must be a numbering method that makes any
There is exactly one difference between the numbered 2 codes of the two vertices that have edges connected to each other.
in 2 and 3 dimensional space This conclusion can be such an image of understanding:
for 2-dimensional space, we just put this square in the first quadrant, making the 4 vertices sit
The mark is in counter-clockwise order (0,0), (1,0), ((), (0,1), and then the coordinates as 2-bit 2-binary
number, then the 4 points are numbered 0,1,3,2.
for 3-dimensional space, we can also overlap one vertex of a cube with the origin and make all
The edges are parallel to the axes, then the coordinates of the 8 points are determined, and finally the coordinates in the 3-dimensional space are viewed
into a 3-bit 2 binary number. For d-dimensional spaces, and so on.
now for an N-1 graph of N-point m-bars (numbered from 0 to--) for each point, will wants to
know if this diagram is isomorphic to a hypercube diagram.

Input

The first line contains an integer q indicating that this data contains a total of Q queries.
Next, the Q group asks, the input format for each group of queries is as follows:
The first line consists of two integers N and M,
Next M line, 2 different integer x, y for each line, indicates that there is a non-forward join in the graph
Points for X and y (0 < = x, y < N)
q<=3,n<=32768,m<=1000000

Output

For each query, output one line, the following:
1, if the query given in the diagram is different in any one hypercube, output-1;
2, if isomorphic to a certain hypercube diagram, then please renumber these n points in the graph, and in this
A row outputs N spaces separated by a space, representing the new number of each point in the original image, making renumbering
To meet the conclusions described in the topic.
Note: Each line of the output file either contains only an integer-1, or it should contain a 0
To N-1 this N number of permutations. If there is more than one set of solutions to the output of either.

A hypercube diagram satisfies:

1. Each vertex has the same degree of K

2. A total of 2k vertices, k2k-1 bar edge

3. After marking any side of the two-end label binary means that only 1-bit difference

4. The label of all points adjacent to each point is different from the bits of this dot label

Not in accordance with 1.2. Direct output-1, in accordance with 1.2. The BFS can be labeled first and verify that the label is valid (the label is not repeated and complies with 3.4.)

BFS, either take a point designator 0, its adjacent point label is a power of 2, take the non-marking and the point adjacent to the labeled Point, marking it as the label of the adjacent labeled point of the bitwise OR value

#include <cstdio>inlineintinput () {intx=0, c=GetChar ();  while(c> $|| c< -) c=GetChar ();  while(c> -&&c< -) x=x*Ten+c- -, c=GetChar (); returnx;}Const intn=32768;intId[n],rs[n],ed[n];inte[ -][n],p[n],dc[n+1];intQ[n],ql=0, qr=0;intn,m,a,b;voidchk () {n=input (); M=input (); BOOLun=0; QL=qr=0; intD=Dc[n]; if(n==1&&m==0) {puts ("0"); return; }    if(! d| | m*2!=N*D) un=1; if(!un) for(intI=0; i<n;i++) p[i]=ed[i]=id[i]=rs[i]=0;  while(m--) {a=input (), b=input (); if(p[a]>=d| | P[b]>=d) un=1; if(un)Continue; E[p[a]++][a]=b; E[P[B]++][b]=A; } ed[0]=1; if(!un) for(intI=0; i<p[0];i++){        intw=e[i][0]; if(Ed[w]) {un=1; Break;} ED[W]=1; Q[QR++]=W; ID[W]=1<<i; }    if(!un) while(ql<qr) {        intw=q[ql++]; ED[W]=1;  for(intI=0; i<p[w];i++){            intu=E[i][w]; if(ed[u]==1)Continue; Id[u]|=Id[w]; if(!Ed[u]) {Ed[u]=2; Q[QR++]=u; }        }    }    if(!un) for(intI=0; i<n;i++){        if(Rs[id[i]]) {un=1; Break;} Rs[id[i]]=i; }    if(!un) for(intI=0; i<n&&!un;i++){        ints=0;  for(intj=0; j<p[i];j++){            intu=E[j][i]; intc=id[u]^Id[i]; if(s&c) {un=1; Break;} if(c!= (C&AMP;-C)) {un=1; Break;} S|=C; }        if(s+1!=n) un=1; }    if(un) puts ("-1"); Else{printf ("%d", id[0]);  for(intI=1; i<n;i++) printf ("%d", Id[i]); Putchar (Ten); }}intMain () { for(intI=0;i< -; i++) dc[1<<i]=i; intt=input ();  while(t--) chk (); return 0;}

bzoj4466 Super Cube