洛谷$P1345\ [USACO5.4]$ 奶牛的电信$Telecowmunication$ 网络流

正解:最小割

解题报告:

传送门$QwQ$

$QwQ$好久没做网络流了来复健下.

这个一看就很最小割趴?考虑咋建图?就把点拆成边权为$1$的边,然后原有的边因为不能割所以边权为$inf$.

然后跑个最小割就做完了,$QwQ$

(虽然但是网络流为啥是蓝的鸭$QwQ$

#include<bits/stdc++.h>
using namespace std;
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define rp(i,x,y) for(int i=x;i<=y;++i)
#define e(i,x) for(int i=head[x];~i;i=edge[i].nxt)
#define E(i,x) for(int &i=cur[x];~i;i=edge[i].nxt)

const int N=+,inf=1e9;
int n,m,s,t,S,T,head[N],cur[N],ed_cnt=-,fr[N],to[N],dep[N];
struct ed{int to,nxt,wei;}edge[N*];
queue<int>Q;

void ad(int x,int y,int z){/*printf("%d->%d:%d\n",y,x,z);*/edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],};head[x]=ed_cnt;}
bool bfs()
{
    memset(dep,,sizeof(dep));dep[S]=;Q.push(S);
    while(!Q.empty()){int nw=Q.front();Q.pop();e(i,nw)if(w(i) && !dep[t(i)])dep[t(i)]=dep[nw]+,Q.push(t(i));}return dep[T];
}
int dfs(int nw,int fl)
{
    if(!(nw^T) || !fl)return fl;;int ret=;
    E(i,nw)if(dep[t(i)]==dep[nw]+){int d=dfs(t(i),min(fl,w(i)));w(i)-=d,w(i^)+=d,fl-=d,ret+=d;if(!fl)return ret;}
    return ret;
}
int dinic(){int ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}

int main()
{
    scanf("%d%d%d%d",&n,&m,&s,&t);memset(head,-,sizeof(head));S=,T=n<<|;
    rp(i,,m)
    {
        int x,y;scanf("%d%d",&x,&y);
        if(x^s && x^t && y^s && y^t)ad(y,x+n,inf),ad(x,y+n,inf);
        else if(!(x^s))ad(y,S,inf);else if(!(y^s))ad(x,S,inf);else if(!(x^t))ad(T,y+n,inf);else ad(T,x+n,inf);
    }
    rp(i,,n)if(i^S && i^T)ad(i+n,i,);;printf("%d\n",dinic());
    return ;
}