BZOJ4205: Card Pairing

For two cards, you can match if there are two property values that are not coprime.

Consider only the prime number within 200, a total of 46, you can create a new 3*46*46 point to represent a class of attribute values that have both of these prime cards.

Then for each card, enumerate its quality factor, up to only 3, so the map to find the maximum flow.

#include <cstdio>const int n=66500,inf=~0u>>2,p=201;struct edge{int T,f;edge*nxt,*pair;} *g[n],*d[n],pool[3500000],*cur=pool;int N,m,i,j,x,y,z,s,t,h[n],gap[n],maxflow;int V[P],fac[P][4],cnt,id12[P][P], id13[p][p],id23[p][p],all;inline int min (int x,int y) {return x<y?x:y;}  inline void Add (int s,int t) {edge*p=cur++;p->t=t;p->f=1;p->nxt=g[s];g[s]=p;  p=cur++;p->t=s;p->f=0;p->nxt=g[t];g[t]=p; G[s]->pair=g[t];g[t]->pair=g[s];}  int sap (int v,int flow) {if (v==t) return flow;  int rec=0; For (Edge*p=d[v];p; p=p->nxt) if (h[v]==h[p->t]+1&&p->f) {int Ret=sap (p->t,min (flow-rec,p->f)    );    p->f-=ret;p->pair->f+=ret;d[v]=p;  if ((Rec+=ret) ==flow) return flow; } if (! (  --GAP[H[V]]) h[s]=t;  gap[++h[v]]++;d [V]=g[v]; Return rec;}  int main () {for (i=2;i<p;i++) if (!v[i]) for (cnt++,j=i;j<p;j+=i) v[j]=1,fac[j][++fac[j][0]]=cnt;  for (i=1;i<=cnt;i++) for (j=1;j<=cnt;j++) Id12[i][j]=++all,id13[i][j]=++all,id23[i][j]=++all; scanf ("%d%d", &n,&m);  s=all+n+m+1,t=s+1;    while (n--) {Add (S,++all);    scanf ("%d%d%d", &x,&y,&z);    for (i=1;i<=fac[x][0];i++) for (j=1;j<=fac[y][0];j++) Add (All,id12[fac[x][i]][fac[y][j]]);    for (i=1;i<=fac[x][0];i++) for (j=1;j<=fac[z][0];j++) Add (All,id13[fac[x][i]][fac[z][j]]);  for (i=1;i<=fac[y][0];i++) for (j=1;j<=fac[z][0];j++) Add (All,id23[fac[y][i]][fac[z][j]]);    } while (m--) {Add (++all,t);    scanf ("%d%d%d", &x,&y,&z);    for (i=1;i<=fac[x][0];i++) for (j=1;j<=fac[y][0];j++) Add (Id12[fac[x][i]][fac[y][j]],all);    for (i=1;i<=fac[x][0];i++) for (j=1;j<=fac[z][0];j++) Add (Id13[fac[x][i]][fac[z][j]],all);  for (i=1;i<=fac[y][0];i++) for (j=1;j<=fac[z][0];j++) Add (Id23[fac[y][i]][fac[z][j]],all);  } for (gap[maxflow=0]=t,i=1;i<=t;i++) d[i]=g[i];  while (h[s]<t) Maxflow+=sap (S,inf); Return printf ("%d", Maxflow), 0;}

　　

BZOJ4205: Card Pairing