BZOJ3275. Number

【题意】

 

 【分析】

和BZOJ3158. 千钧一发基本一致

【代码】

#include<bits/stdc++.h>
using namespace std;
#define mp make_pair
#define fi first
#define se second
#define lson now<<1
#define rson now<<1|1
typedef long long ll;
const int maxn=3005;
const int maxm=1000005;
int n,m,a[maxn];
int S,T;
const ll inf=1e17;
int head[10005],tot=1,cur[10005];
struct edge
{
    int to,nxt;
    ll v;
}e[maxm<<1];
void add(int x,int y,ll z)
{
    e[++tot].to=y; e[tot].nxt=head[x]; e[tot].v=z; head[x]=tot;
    e[++tot].to=x; e[tot].nxt=head[y]; e[tot].v=0; head[y]=tot;
}
int dep[10005];
bool bfs()
{
    for(int i=S;i<=T;i++)
        dep[i]=-1,cur[i]=head[i];
    // memset(dep,-1,sizeof(dep));
    // memcpy(cur,head,sizeof(cur));
    queue <int> q;
    dep[S]=0;
    q.push(S);
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        for(int i=head[u];i;i=e[i].nxt)
        {
            int to=e[i].to;
            if(dep[to]!=-1 || !e[i].v) continue;
            q.push(to);
            dep[to]=dep[u]+1;
        }
    }
    return (dep[T]!=-1);
}
ll dfs(int u,ll flow)
{
    if(u==T) return flow;
    ll res=0;
    for(int &i=cur[u];i;i=e[i].nxt)
    {
        int to=e[i].to;
        if(dep[to]!=dep[u]+1 || e[i].v<=0) continue;
        ll tmp=dfs(to,min(e[i].v,flow));
        flow-=tmp; res+=tmp;
        e[i].v-=tmp; e[i^1].v+=tmp;
        if(!flow) break;
    }
    if(!res) dep[u]=-1;
    return res;
}
ll ans;
ll dinic()
{
    ll ans=0;
    while(bfs())
    {
        ans+=dfs(S,inf);
    }
    return ans;
}
int gcd(int x,int y)
{
    if(!y) return x;
    return gcd(y,x%y); 
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    S=0; T=n*2+1;
    ll sum=0;
    for(int i=1;i<=n;i++)
        sum+=a[i],add(S,i,a[i]),add(i+n,T,a[i]);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            if(gcd(a[i],a[j])>1) continue;
            int k=sqrt(1LL*a[i]*a[i]+1LL*a[j]*a[j]);
            if(1LL*k*k!=1LL*a[i]*a[i]+1LL*a[j]*a[j]) continue;
            add(i,j+n,inf);
        }
    printf("%lld",sum-dinic()/2);
    return 0;
}