bzoj 3158: 千钧一发 （最小割）

题目描述

传送门

题目大意：给出n个数，每个数有两权a,b。
两个数能同时选，必须满足下面至少一个条件
(1) gcd(ai,aj)!=1
(2)不存在整数T满足 ai2+aj2=T2
求所选集合 ∑b 的最大值

题解

设源汇分别为S,T,对于每个数拆成两个点xi,yi
S->xi 容量为bi
yi->T 容量为bi
xi->yi 容量为inf，两点不满足上述条件。
跑最大流求最小割即可。

代码

#include  
#include  
#include  
#include  
#include  
#include  
#define N 4000003  
#define inf 1000000000  
#define LL long long 
using namespace std;  
int remain[N],point[N],nxt[N],v[N],n,tot,S,T;  
int deep[N],last[N],num[N],cur[N],a[N],b[N];  
void add(int x,int y,int z)  
{  
    tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=z;  
    tot++; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;  
    //cout<
}  
long long pow(long long x){  
    return x*x;  
}  
int gcd(int x,int y)  
{  
    int r;  
    while (y) {  
        r=x%y;  
        x=y; y=r;  
    }  
    return x;  
}  
bool check(int i,int j)  
{  
    if (gcd(a[i],a[j])!=1) return true;  
    long long t=pow(a[i])+pow(a[j]);  
    long long t1=sqrt(t);  
    if (t1*t1!=t) return true;  
    return false;  
}  
int addflow(int s,int t)  
{  
    int now=t; int ans=inf;  
    while (now!=s) {  
        ans=min(ans,remain[last[now]]);  
        now=v[last[now]^1];  
    }  
    now=t;  
    while (now!=s) {  
        remain[last[now]]-=ans;  
        remain[last[now]^1]+=ans;  
        now=v[last[now]^1];  
    }  
    return ans;  
}  
void bfs(int s,int t)  
{  
   for (int i=s;i<=t;i++) deep[i]=t;  
   deep[t]=0;   
   queue<int> p; p.push(t);  
   while (!p.empty()) {  
     int now=p.front(); p.pop();  
     for (int i=point[now];i!=-1;i=nxt[i])  
      if (deep[v[i]]==t&&remain[i^1])  
       deep[v[i]]=deep[now]+1,p.push(v[i]);  
   }  
}  
int isap(int s,int t)  
{  
    int ans=0; bfs(s,t);  
    for (int i=1;i<=t;i++) cur[i]=point[i],num[deep[i]]++;  
    int now=s;  
    while (deep[s]if (now==t) {  
            ans+=addflow(s,t);  
            now=s;  
        }  
        bool pd=false;  
        for (int i=cur[now];i!=-1;i=nxt[i])  
         if (deep[now]==deep[v[i]]+1&&remain[i]) {  
            pd=true;  
            last[v[i]]=i; cur[now]=i;  
            now=v[i]; break;  
         }  
        if (!pd) {  
            int minn=t;  
            for (int i=point[now];i!=-1;i=nxt[i])  
             if (remain[i]) minn=min(deep[v[i]],minn);  
            if (!--num[deep[now]]) break;  
            num[deep[now]=minn+1]++;  
            cur[now]=point[now];  
            if (now!=s) now=v[last[now]^1];  
        }  
    }  
    return ans;  
} 
int main()  
{  
    freopen("a.in","r",stdin);  
    freopen("my.out","w",stdout);
    scanf("%d",&n);  
    S=1; T=2*n+2;  
    tot=-1;
    memset(point,-1,sizeof(point));
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);  
    int sum=0;  
    for (int i=1;i<=n;i++) {  
        scanf("%d",&b[i]); sum+=b[i];  
        add(S,i+1,b[i]);  
        add(i+n+1,T,b[i]);  
    }  
    for (int i=1;i<=n;i++)  
     for (int j=1;j<=n;j++)  
      if (!check(i,j)) add(i+1,j+n+1,inf);  
    printf("%d\n",sum-isap(S,T)/2);  
}