【bzoj2502】清理雪道 有上下界的网络流

发现了一种不错的最小流求法。

有源汇的最小流
源点S向每个点连一条容量为(0,inf)的边
每个点向汇点T连一条容量为(0,inf)的边
原图的每条边变成容量为(1,inf)的边
最小流求法：
二分一个答案，每次在S到T间连一条容量为(0,x)的边
判断是否可行

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#define maxn 210#define maxm 100010#define inf 1000000000using namespace std;int head[maxn],to[maxm],c[maxm],next[maxm],q[maxn],d[maxn];int a[maxn][maxn];int in[maxn],cnt[maxn];int s,t,n,m,num,S,T,ans;void addedge(int x,int y,int z){num++;to[num]=y;c[num]=z;next[num]=head[x];head[x]=num;num++;to[num]=x;c[num]=0;next[num]=head[y];head[y]=num;}bool bfs(){memset(d,-1,sizeof(d));int l=0,r=1;q[1]=S;d[S]=0;while (l<r){int x=q[++l];for (int p=head[x];p;p=next[p])  if (c[p] && d[to[p]]==-1)  {  d[to[p]]=d[x]+1;  q[++r]=to[p];  }}if (d[T]==-1) return 0; else return 1;}int find(int x,int low){if (x==T || low==0) return low;int totflow=0;for (int p=head[x];p;p=next[p])  if (c[p] && d[to[p]]==d[x]+1)  {  int a=find(to[p],min(low,c[p]));  c[p]-=a;c[p^1]+=a;  low-=a;totflow+=a;  if (low==0) return totflow;  }if (low) d[x]=-1;return totflow;}void Dinic(){int ans=0;while (bfs()) ans+=find(S,inf);}bool check(int x){num=1;s=0;t=n+1;S=n+2;T=n+3;memset(in,0,sizeof(in));memset(head,0,sizeof(head));memset(cnt,0,sizeof(cnt));addedge(t,s,x);for (int i=1;i<=n;i++)  for (int j=1;j<=n;j++)    if (a[i][j])    {    in[i]--;in[j]++;    addedge(i,j,inf);    }for (int i=1;i<=n;i++) addedge(s,i,inf),addedge(i,t,inf);for (int i=1;i<=n;i++)  if (in[i]>0) cnt[i]=num+1,addedge(S,i,in[i]);  else if (in[i]<0) cnt[i]=num+1,addedge(i,T,-in[i]);Dinic();for (int i=1;i<=n;i++)  if (cnt[i] && c[cnt[i]]) return 0;return 1;}int main(){scanf("%d",&n);for (int i=1;i<=n;i++){int x;scanf("%d",&x);for (int j=1;j<=x;j++){int x;scanf("%d",&x);a[i][x]=1;}}int l=0,r=n*n,ans;while (l<=r){int mid=(l+r)/2;if (check(mid)) ans=mid,r=mid-1; else l=mid+1;}printf("%d\n",ans);return 0;}