Bzoj 1305 Dance Dance

Network flow.

First, the answer to the question is whether the X-ray dance is feasible.

Consider building a map, set three points for each person, representing the whole, like and dislike.

The source point connects a single edge of X to each boy's point.

The whole point of each boy likes to point to a side with a capacity of positive infinity and a side with a capacity of K to dislike points.

Every boy likes to point to all the girls he likes. A side that has a capacity of one, does not like to point to all the girls he doesn't like. Don't like to point to a one-size side.

Each girl likes to point to the whole point with a positive infinity edge, and does not like to point the whole point to a single edge with a capacity of K.

Each girl's whole point to a meeting point is an X-size edge.

Check is the determination of whether the maximum stream equals n*x.

In this case, the definition of flow is the number of dance, the whole point to the Yuanhui point of the edge to ensure that each round is exactly paired with n pairs, such as a round in the presence of a female student corresponding to multiple boys, the final maximum flow will be reduced.

#include <cstring>#include<cstdio>#include<algorithm>#include<iostream>#include<queue>using namespacestd;Const intdian=305;Const intbian=40005;Const intinf=0x3f3f3f3f;intN,k,tot;ints,t;intH[dian],nxt[bian],ver[bian],val[bian],ch[dian];Charmap[ -][ -];voidAddintCanintBbintcc) {Tot++;ver[tot]=bb;val[tot]=cc;nxt[tot]=h[aa];h[aa]=tot; Tot++;ver[tot]=aa;val[tot]=0; nxt[tot]=h[bb];h[bb]=tot;}intBhintAaintPintOK) {    return(aa-1)*3+p+ok*3*N;}BOOLTell () {memset (ch,-1,sizeof(CH)); Queue<int>Q;    Q.push (S); Ch[s]=0;  while(!Q.empty ()) {        intt=Q.front ();        Q.pop ();  for(intI=h[t];i;i=Nxt[i])if(ch[ver[i]]==-1&&Val[i])                {Q.push (ver[i]); Ch[ver[i]]=ch[t]+1; }    }    returnch[t]!=-1;}intZengintAintb) {    if(a==T)returnb; intR=0;  for(intI=h[a];i&&b>r;i=Nxt[i])if(ch[ver[i]]==ch[a]+1&&Val[i]) {            intT=zeng (Ver[i],min (b-r,val[i])); Val[i]-=t,r+=t,val[i^1]+=T; }    if(!r) Ch[a]=-1; returnR;}intDinic () {intR=0, T;  while(Tell ()) while(t=Zeng (S,inf)) R+=T; returnR;}BOOLCheckintmid) {memset (NXT,0,sizeof(NXT)); memset (H,0,sizeof(h)); Tot=1;  for(intI=1; i<=n;i++) {Add (S,BH (i,1,0), mid); Add (BH (i,1,1), T,mid); Add (BH (i,1,0), BH (i,2,0), INF); Add (BH (i,1,0), BH (i,3,0), k); Add (BH (i,2,1), BH (i,1,1), INF); Add (BH (i,3,1), BH (i,1,1), k); }     for(intI=1; i<=n;i++)         for(intj=1; j<=n;j++){            if(map[i][j]=='Y') Add (BH (i,2,0), BH (J,2,1),1); ElseAdd (BH (i,3,0), BH (J,3,1),1); }    if(Dinic () ==n*mid)return true; return false;}intBinintLintR) {    intmid;  while(l<R) {Mid= (l+r+1) >>1; if(check (mid)) L=mid; ElseR=mid-1; }    returnl;}intMain () {scanf ("%d%d",&n,&k); S=6*n+1, t=6*n+2;  for(intI=1; i<=n;i++) scanf ("%s", map[i]+1); printf ("%d", Bin (0, -)); return 0;}

Bzoj 1305 Dance Dance