bzoj3275: Number

1A了，最小割233，拆點，s向i1連a[i]的邊，i2向t連a[i]的邊，有限制的i1與j2,j1與i2連inf的邊，答案就是sum-ans/2啦！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
queue<int> q;
const int inf=0x7f7f7f7f;
int read()
{
    char ch=getchar();int f=0;
    while(ch<'0'||ch>'9')ch=getchar();
    while(ch>='0'&&ch<='9'){f=(f<<1)+(f<<3)+ch-'0';ch=getchar();}
    return f;
}
int gcd(int x,int y)
{
    return !y?x:gcd(y,x%y);
}
bool check(int x,int y)
{
    if(gcd(x,y)!=1) return 0;
    int z=x*x+y*y;
    int temp=sqrt(z);
    if(temp*temp!=z) return 0; 
    return 1;
}
struct node
{
    int from;
    int to;
    int next;
    int w;
}edge[5000005];
int tot,head[10005],n,a[100005],s,t,ans,dis[10005];
void add(int u,int v,int w)
{
    edge[tot].from=u;
    edge[tot].to=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}
bool bfs()
{
    memset(dis,-1,sizeof(dis));
    q.push(s);
    dis[s]=0;
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        for(int i=head[x];i!=-1;i=edge[i].next)
        {
            if(dis[edge[i].to]==-1&&edge[i].w)
            {
                dis[edge[i].to]=dis[x]+1;
                q.push(edge[i].to);
            }
        }
    }
    if(dis[t]==-1) return 0;
    return 1;
}
int dfs(int x,int flow)
{
    int ret=0;
    if(x==t||!flow) return flow;
    for(int i=head[x];i!=-1;i=edge[i].next)
    {
        if(dis[edge[i].to]==dis[x]+1)
        {
            int f=dfs(edge[i].to,min(edge[i].w,flow));
            edge[i].w-=f;edge[i^1].w+=f;
            ret+=f;flow-=f;
            if(!flow) break;
        }
    }
    if(!ret) dis[x]=-1;
    return ret;
}
void dinic()
{
    while(bfs()) ans+=dfs(s,0x7f7f7f7f);
}
int main()
{
    memset(head,-1,sizeof(head));
    n=read();
    s=0,t=2*n+1;int sum=0;
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        sum+=a[i];
    }
    for(int i=1;i<=n;i++)
    {
        add(s,i,a[i]);
        add(i,s,0);
        add(i+n,t,a[i]);
        add(t,i+n,0);
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(check(a[i],a[j]))
            {
                add(i,j+n,inf);
                add(j+n,i,0);
                add(j,i+n,inf);
                add(i+n,j,0);
            }
        }
    }
    dinic();
    cout<<sum-ans/2;
}