ZOJ 1015 Fishing Net (judging the chord graph)

Title Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=15

Test instructions: given a graph. Judging is not a chord chart?

Idea: (1) God horse is a string diagram? For an image without a direction, if any one of the rings with a length greater than 3 has an edge connected to a two-point that is not adjacent to the ring, then this figure is called a chord graph.

(2) What is a regiment? A group is a sub-image of the original, and the sub-map contains some of the points of the original, so it is necessary to include the edges between those points. And the regiment is not a normal sub-figure but a complete sub-graph, that is, there are edges between any two vertices of the sub-graph. The following ABCD is a regiment of the original.

(3) Perfect elimination sequence: the sequence of a point of the original image (each point appears and happens exactly once) V1, v2,......, vn to meet {VI, vi+1,..., vn} The composition of the sub-diagram for the regiment.

(4) An no-show graph is a chord graph when and only if it has a perfect elimination sequence.

(5) How to calculate the perfect elimination sequence? The most popular algorithm: the order of the points from N to 1 sequentially (the point labeled I appears in the perfect elimination sequence of the first I). Set Label[i] indicates that the first point is adjacent to how many labeled points, each time you select Label[i] The largest unlabeled point to label. Note that the perfect elimination sequence is only calculated here, but it has not been decided after this is a chord graph.

(6) How to determine the original image from the perfect elimination sequence is not a chord chart? The simplest way is to determine, in turn, whether all the points adjacent to VI in the {vi+1,..., vn} constitute a regiment. This can be optimized by setting {vi+1,..., vn} where all the points adjacent to VI are VJ1,......, VJK. Simply determine if the vj1 is adjacent to the Vj2,......, vjk.

int n,m,g[n][n];int D[n],a[n],h[n],p[n];int OK () {    int i,j,u;    Vector<int> V;    FORL1 (i,n)    {        v.clear ();        FOR1 (J,n) if (G[a[i]][j]) if (p[j]>i) V.PB (j);        For (J=1;j<sz (V); j + +) if (P[v[0]]>p[v[j]])        {            swap (v[0],v[j]);        }        For (J=1;j<sz (V); j + +)        {            if (!g[v[0]][v[j]) return 0;}    }    return 1;} int main () {    Rush (n)    {        RD (m);        if (!n&&!m) break;        int i,j,k,u,v;        CLR (g,0);        FOR1 (i,m)        {            RD (u,v);            g[u][v]=g[v][u]=1;        }        CLR (d,0); CLR (h,0);        FORL1 (i,n)        {            u=-1;            FOR1 (J,n) if (!h[j]&&d[j]>u) u=d[j],k=j;            A[i]=k; H[k]=1; p[k]=i;            FOR1 (J,n) if (g[k][j]) d[j]++;        }        if (ok ())) puts ("Perfect");        Else puts ("imperfect");        Puts ("");    }    return 0;}