跑最大费用最大流，注意到每次 spfa 出来的 cost 一定是越来越少的，啥时小于 (0) 了就停了吧。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
typedef long long ll;
int n, a[205], b[205], c[205], dd[205], hea[205], cnt, ss, tt, pre[205];
const int oo=0x3f3f3f3f;
bool vis[205];
ll dis[205], ans;
queue<int> d;
struct Edge{
int too, nxt, val;
ll cst;
}edge[200005];
int calc(int x){
int re=0;
for(int i=2; i*i<=x; i++)
if(x%i==0){
while(x%i==0){
x /= i;
re++;
}
}
if(x!=1) re++;
return re;
}
void add_edge(int fro, int too, int val, ll cst){
edge[cnt].nxt = hea[fro];
edge[cnt].too = too;
edge[cnt].val = val;
edge[cnt].cst = cst;
hea[fro] = cnt++;
}
void addEdge(int fro, int too, int val, ll cst){
add_edge(fro, too, val, cst);
add_edge(too, fro, 0, -cst);
}
bool spfa(){
memset(dis, 0x3f, sizeof(dis));
memset(pre, -1, sizeof(pre));
d.push(ss);
dis[ss] = 0;
vis[ss] = true;
while(!d.empty()){
int x=d.front();
d.pop();
vis[x] = false;
for(int i=hea[x]; i!=-1; i=edge[i].nxt){
int t=edge[i].too;
if(dis[t]>dis[x]+edge[i].cst && edge[i].val>0){
dis[t] = dis[x] + edge[i].cst;
pre[t] = i;
if(!vis[t]){
vis[t] = true;
d.push(t);
}
}
}
}
return dis[tt]!=0x3f3f3f3f3f3f3f3f;
}
void mcmf(){
ll cost=0;
while(spfa()){ int tmp=oo;
for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too])
tmp = min(tmp, edge[i].val);
if(cost+(ll)tmp*dis[tt]<=0){
cost += (ll)tmp * dis[tt];
ans += tmp;
for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too]){
edge[i].val -= tmp;
edge[i^1].val += tmp;
}
}
else{
ans += cost / (-dis[tt]);
return ;
}
}
}
int main(){
memset(hea, -1, sizeof(hea));
cin>>n;
for(int i=1; i<=n; i++) scanf("%d", &a[i]);
for(int i=1; i<=n; i++) scanf("%d", &b[i]);
for(int i=1; i<=n; i++) scanf("%d", &c[i]);
for(int i=1; i<=n; i++) dd[i] = calc(a[i]);
ss = 0; tt = n + 1;
for(int i=1; i<=n; i++){
if(dd[i]&1){
addEdge(ss, i, b[i], 0);
for(int j=1; j<=n; j++)
if(a[j]%a[i]==0 && dd[j]==dd[i]+1) addEdge(i, j, oo, (ll)-c[i]*c[j]);
else if(a[i]%a[j]==0 && dd[i]==dd[j]+1) addEdge(i, j, oo, (ll)-c[i]*c[j]);
}
else addEdge(i, tt, b[i], 0);
}
mcmf();
printf("%lld
", ans);
return 0;
}