【bzoj4514】: [Sdoi2016]数字配对 图论-费用流

【bzoj4514】: [Sdoi2016]数字配对

好像正常的做法是建二分图？

我的是拆点然后

S->i cap=b[i] cost=0

i'->T cap=b[i] cost=0

然后能匹配的两点i,j 连 i->j' cap=inf cost=c[i]*c[j]

跑最大费用流，直到 cost<0 或 全部增广完

最后flow/2就是答案

 /* http://www.cnblogs.com/karl07/ */
 #include <cstdlib>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <queue>
 using namespace std;

 #define ll long long
 const ll inf=1e18;
 const int N=1e5+;
 struct edge{
     int from,next,to;
     ll v,c;
 }e[N];
 int first[N],pr[N],prime[N],inq[N],lst[N];
 ll A[N],B[N],C[N],dis[N],minf[N];
 int S=,T=,ade=,P,n;
 queue <int> Q;

 void addedge(int x,int y,ll v,ll c){
     e[++ade].to=y;
     e[ade].from=x;
     e[ade].next=first[x];
     e[ade].v=v;
     e[ade].c=c;
     first[x]=ade;
 }

 void ADE(int x,int y,ll v,ll c){
     addedge(x,y,v,c);
     addedge(y,x,-v,);
 }

 void Prime(){
     for (int i=;i<;i++) prime[i]=;
     for (int i=;i<;i++){
         if (prime[i]){
             pr[++pr[]]=i;
             for (int j=i+i;j<;j+=i) prime[j]=;
         }
     }
 }

 bool check(ll x){
     if (x==) return ;
     for (int i=;i<=pr[] && pr[i]<x ;i++) if (!(x%pr[i])) return ;
     return ;
 }

 #define s e[x].to
 #define v e[x].v
 #define cap e[x].c
 #define Cap e[x^1].c
 bool SPFA(ll &mf,ll &mc){
     for (int i=;i<=n*+;i++) dis[i]=-inf,minf[i]=inf;
     Q.push(S),inq[S]=,dis[S]=;
     while (!Q.empty()){
         int p=Q.front();
         Q.pop(),inq[p]=;
         for (int x=first[p];x;x=e[x].next){
             if (dis[s]<dis[p]+v && cap>){
                 dis[s]=dis[p]+v;
                 lst[s]=x;
                 minf[s]=min(minf[p],cap);
                 if (!inq[s]) Q.push(s),inq[s]=;
             }
         }
     }
     if (dis[T]==-inf) return ;
     for (int x=lst[T];x;x=lst[e[x].from]) {cap-=minf[T],Cap+=minf[T];}
     mf+=minf[T];
     mc+=dis[T]*minf[T];
     if (mc<){
         mf-=mc/dis[T]+(mc%dis[T]!=);
         return ;
     }
     return ;
 }

 void mcmf(){
     ll mf=,mc=;
     while (SPFA(mf,mc));
     printf("%lld\n",mf/);
 }
 #undef s
 #undef v
 #undef c
 #undef C

 int main(){
     Prime();
     scanf("%d",&n);
     for (int i=;i<=n;i++) scanf("%lld",&A[i]);
     for (int i=;i<=n;i++) scanf("%lld",&B[i]);
     for (int i=;i<=n;i++) scanf("%lld",&C[i]);
     for (int i=;i<=n;i++){
         ADE(S,i+,,B[i]);
         ADE(i+n+,T,,B[i]);
         for (int j=;j<=n;j++){
             if (A[i]>A[j] && A[i]%A[j]==){
                 if (check(A[i]/A[j])){
                     ADE(i+,j+n+,C[i]*C[j],inf);
                     ADE(j+,i+n+,C[i]*C[j],inf);
                 }
             }
         }
     }
     mcmf();
     return ;
 }