BZOJ3158 千钧一发（最小割）

　　可以看做一些物品中某些互相排斥求最大价值。如果这是个二分图的话，就很容易用最小割了。

　　观察其给出的条件间是否有什么联系。如果两个数都是偶数，显然满足条件二；而若都是奇数，则满足条件一，因为式子列出来发现一定不能写成完全平方数。那么这就是个二分图了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 1010
#define inf 1000000000
#define S 0
#define T 1001
int n,p[N],a[N],b[N],ans=0,t=-1;
int d[N],cur[N],q[N];
struct data{int to,nxt,cap,flow;
}edge[N*N<<1];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,p[y]=t;
}
bool bfs()
{
    memset(d,255,sizeof(d));d[S]=0;
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
        {
            d[edge[i].to]=d[x]+1;
            q[++tail]=edge[i].to;
        }
    }while (head<tail);
    return ~d[T];
}
int work(int k,int f)
{
    if (k==T) return f;
    int used=0;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+1==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
        edge[i].flow+=w,edge[i^1].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==0) d[k]=-1;
    return used;
}
void dinic()
{
    while (bfs())
    {
        memcpy(cur,p,sizeof(p));
        ans-=work(S,inf);
    }
}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3158.in","r",stdin);
    freopen("bzoj3158.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++) ans+=b[i]=read();
    memset(p,255,sizeof(p));
    for (int i=1;i<=n;i++)
    if (a[i]&1)
        for (int j=1;j<=n;j++)
        if (!(a[j]&1)&&((long long)sqrt(1ll*a[i]*a[i]+1ll*a[j]*a[j]))*((long long)sqrt(1ll*a[i]*a[i]+1ll*a[j]*a[j]))==1ll*a[i]*a[i]+1ll*a[j]*a[j]&&gcd(a[i],a[j])==1)
        addedge(i,j,inf);
    for (int i=1;i<=n;i++)
    if (a[i]&1) addedge(S,i,b[i]);
    else addedge(i,T,b[i]);
    dinic();
    cout<<ans;
    return 0;
}