$loj\ 2031\ [SDOI2016]$数字配对 网络流

正解:网络流

解题报告:

我永远喜欢$loj$!

显然先预处理哪些$a$之间可以连边,然后考虑建两排点,连流量为$c_{i}\cdot c_{j}$,然后$ST$连$inf$,跑个费用流?

然后现在碰到了两个问题$QwQ$

第一个说不能重复使用?

可以考虑这样儿,就如果$(i,j)$是一对可行点对,那就要连$(i,{j}'),(j,{i}')$,显然不可能只跑一条嘛,然后最后答案除以2就成$QwQ$

第二个说问在费用大于等于0的前提下的最大流,,,就很单纯不做作和之前的题都不太一样,,,

考虑贪心,首先说费用最大先取反呗.

然后就每次增广的时候判断下,如果之前增广的费用+当前增广费用>0就取之前算出来的最大流量就成$QwQ$

正确性在于每次跑的都最短路?所以就十分显然不说了$QwQ$

$over$

昂对了记得开$ll$,,,$100pts->50pts$还是有点难受的,,,$QAQ$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define int long long
#define t(i) edge[i].to
#define n(i) edge[i].nxt
#define w(i) edge[i].wei
#define fy(i) edge[i].fy
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define e(i,x) for(ri i=head[x];~i;i=n(i))

const int N=+,M=+,inf=1e9;
int head[N],ed_cnt=-,S,T,dis[N],fr_nod[N],fr_ed[N],cost,n,a[N],c[N],fl;
bool vis[N];
struct ed{int to,nxt,wei,fy;}edge[M<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il void ad(ri x,ri y,ri z,ri p)
{
    //printf("%d -> %d : %d , %d\n",y,x,z,p);
    edge[++ed_cnt]=(ed){x,head[y],z,p};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],,-p};head[x]=ed_cnt;}
il bool spfa()
{
    queue<int>Q;Q.push(S);memset(dis,,sizeof(dis));dis[S]=;vis[S]=;
    while(!Q.empty())
    {
        ri nw=Q.front();Q.pop();vis[nw]=;//printf("nw=%d\n",nw);
        e(i,nw)
        {
            if(w(i) && fy(i)+dis[nw]<dis[t(i)])
            {dis[t(i)]=dis[nw]+fy(i);fr_nod[t(i)]=nw;fr_ed[t(i)]=i;if(!vis[t(i)])Q.push(t(i)),vis[t(i)]=;}
        }
    }
    if(dis[T]==dis[T+])return ;
    ri flow=dis[T+];
    for(ri i=T;i!=S;i=fr_nod[i])flow=min(flow,w(fr_ed[i]));
    for(ri i=T;i!=S;i=fr_nod[i])w(fr_ed[i])-=flow,w(fr_ed[i]^)+=flow;
    cost+=flow*dis[T];if(cost>){cost-=flow*dis[T];fl+=1ll*cost/(-dis[T]);return ;}fl+=flow;return ;
}
il bool check(ri x){for(ri i=;i*i<=x;++i)if(!(x%i))return ;return ;}
il void pre()
{
    rp(i,,n)
    {
        rp(j,,n)
        {
            if(i==j)continue;if(a[i]%a[j])continue;if(a[i]==a[j] && i<j)continue;
            if(check(a[i]/a[j]))ad(j+n,i,inf,-c[i]*c[j]),ad(i+n,j,inf,-c[i]*c[j]);
        }
    }
}

 main()
{
    //freopen("4068.in","r",stdin);freopen("4068.out","w",stdout);
    n=read();S=;T=n<<|;memset(head,-,sizeof(head));
    rp(i,,n)a[i]=read();rp(i,,n){ri x=read();ad(i,S,x,);ad(T,i+n,x,);}rp(i,,n)c[i]=read();pre();
    while(spfa());printf("%d\n",fl/);
    return ;
}