【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+5;
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=1,T=2,ade=1,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,0);
}

void Prime(){
    for (int i=2;i<40000;i++) prime[i]=1;
    for (int i=2;i<40000;i++){
        if (prime[i]){
            pr[++pr[0]]=i;
            for (int j=i+i;j<40000;j+=i) prime[j]=0;
        }
    }
}

bool check(ll x){
    if (x==1) return 0;
    for (int i=1;i<=pr[0] && pr[i]<x ;i++) if (!(x%pr[i])) return 0;
    return 1;
}

#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=1;i<=n*2+2;i++) dis[i]=-inf,minf[i]=inf;
    Q.push(S),inq[S]=1,dis[S]=0;
    while (!Q.empty()){
        int p=Q.front();
        Q.pop(),inq[p]=0;
        for (int x=first[p];x;x=e[x].next){
            if (dis[s]<dis[p]+v && cap>0){
                dis[s]=dis[p]+v;
                lst[s]=x;
                minf[s]=min(minf[p],cap);
                if (!inq[s]) Q.push(s),inq[s]=1;
            } 
        }
    }
    if (dis[T]==-inf) return 0;
    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<0){
        mf-=mc/dis[T]+(mc%dis[T]!=0);
        return 0;
    }
    return 1;
}

void mcmf(){
    ll mf=0,mc=0;
    while (SPFA(mf,mc));
    printf("%lld
",mf/2);
}
#undef s
#undef v
#undef c
#undef C


int main(){
    Prime();
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%lld",&A[i]);
    for (int i=1;i<=n;i++) scanf("%lld",&B[i]);
    for (int i=1;i<=n;i++) scanf("%lld",&C[i]);
    for (int i=1;i<=n;i++){
        ADE(S,i+2,0,B[i]);
        ADE(i+n+2,T,0,B[i]);        
        for (int j=1;j<=n;j++){
            if (A[i]>A[j] && A[i]%A[j]==0){
                if (check(A[i]/A[j])){
                    ADE(i+2,j+n+2,C[i]*C[j],inf);
                    ADE(j+2,i+n+2,C[i]*C[j],inf);                    
                }
            }
        }
    }
    mcmf();
    return 0;
}