BZOJ4514 [Sdoi2016]数字配对

把每个点拆成两个分别在左右两边，只要两两之间能发生关系就在左右之间连边，容量为无穷，费用为代价，源连左边每个点容量为数量，费用0，右边每个点连汇容量为数量，费用为零，跑费用流，流到费用为非负的极限，最小费用最大流贪心流是正确的，非负费用最大流贪心流也一定是正确的

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<ctime>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<map>
using namespace std;
#define MAXN 4010
#define MAXM 500010
#define INF 1000000000000000ll
#define MOD 1000000007
#define ll long long
#define eps 1e-8
struct vec{
ll to;
ll fro;
ll f;
ll v;
};
ll n;
ll a[MAXN],b[MAXN],c[MAXN];
vec mp[MAXM];
ll tai[MAXN],cnt=1;
ll q[MAXN],hd,tl;
ll dis[MAXN];
bool iq[MAXN];
ll rd[MAXN];
ll s,t;
ll answ;
ll ansf;
inline void be(ll x,ll y,ll z,ll zz){
mp[++cnt].to=y;
mp[cnt].fro=tai[x];
tai[x]=cnt;
mp[cnt].f=z;
mp[cnt].v=zz;
}
inline void bse(ll x,ll y,ll z,ll zz){
be(x,y,z,zz);
be(y,x,0,-zz);
}
bool spfa(){
ll i,x,y;
for(i=0;i<=t;i++){
dis[i]=-INF;
}
hd=tl=0;
dis[s]=0;
iq[q[tl++]=s]=1;
while(hd!=tl){
iq[x=q[hd++]]=0;
hd%=MAXN;
for(i=tai[x];i;i=mp[i].fro){
y=mp[i].to;
if(mp[i].f&&dis[y]<dis[x]+mp[i].v){
dis[y]=dis[x]+mp[i].v;
rd[y]=i;
if(!iq[y]){
iq[q[tl++]=y]=1;
tl%=MAXN;
}
}
}
}
if(answ+dis[t]<0){
return 0;
}
ll flow=INF;
if(dis[t]<0){
flow=answ/(-dis[t]);
}
for(i=t;i!=s;i=mp[rd[i]^1].to){
//<<i<<'!';
flow=min(flow,(ll)mp[rd[i]].f);
}
//<<endl;
for(i=t;i!=s;i=mp[rd[i]^1].to){
mp[rd[i]].f-=flow;
mp[rd[i]^1].f+=flow;
}
ansf+=flow;
answ+=flow*dis[t];
//<<dis[t]<<'@'<<flow<<endl;
return 1;
}
ll p[MAXM],tot;
bool np[MAXM];
bool jud(int x){
int i;
if(x==1||x==0){
return 0;
}
int y=sqrt(x);
for(i=2;i<=y;i++){
if(!(x%i)){
return 0;
}
}
return 1;
}
int main(){
ll i,j;

scanf("%lld",&n);
s=(n<<1|1)+1,t=s+1;
for(i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
for(i=1;i<=n;i++){
scanf("%lld",&b[i]);
bse(s,i<<1,b[i],0);
bse(i<<1|1,t,b[i],0);
}
for(i=1;i<=n;i++){
scanf("%lld",&c[i]);
}

for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
if(a[j]&&!(a[i]%a[j])&&jud(a[i]/a[j])){
bse(i<<1,j<<1|1,INF,c[i]*c[j]);
}
if(a[i]&&!(a[j]%a[i])&&jud(a[j]/a[i])){
bse(i<<1,j<<1|1,INF,c[i]*c[j]);
}
}
}

//*
while(spfa()){

}
//*/
printf("%lld\n",ansf/2);
return 0;
}

/*

3
2 4 8
3 200 7
5 1 -7

*/