题目描述
传送门
题解
建原图很简单: 对于能到达的点x,y,x->y,[1,inf] s->i,[0,inf];i->t,[0,inf]
将原图进行改造 建立附加源汇ss,tt 对于原图中有的边x->y,[b,c],连边x->y,c-b 记某一个点的权d(i)为所有流入这个点的边的下界和-所有流出这个点的边的下界和 若d(i)>0,连边ss->i,d(i) 若d(i)<0,连边i->tt,-d(i) 然后对ss->tt跑最大流 然后连边t->s,inf 再对ss->tt跑最大流 此时t->s,inf这条边的实际流量就是原图中的最小流
代码
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
#define N 110
#define E 30005
#define inf 1000000000
int n,m,x,s,t,ss,tt,maxflow;
int tot,point[N],nxt[E],v[E],remain[E];
int d[N],deep[N],num[N],last[N],cur[N];
queue <int> q;
void addedge(
int x,
int y,
int cap)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=
0;
}
void bfs(
int t)
{
for (
int i=
1;i<=t;++i) deep[i]=t;
deep[t]=
0;
for (
int i=
1;i<=t;++i) cur[i]=point[i];
while (!q.empty()) q.pop();
q.push(t);
while (!q.empty())
{
int now=q.front();q.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;
q.push(v[i]);
}
}
}
int addflow(
int s,
int t)
{
int now=t,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 isap(
int s,
int t)
{
bfs(t);
for (
int i=
1;i<=t;++i) ++num[deep[i]];
int now=s;
while (deep[s]<t)
{
if (now==t)
{
maxflow+=addflow(s,t);
now=s;
}
bool has_find=
false;
for (
int i=cur[now];i!=-
1;i=nxt[i])
if (deep[v[i]]+
1==deep[now]&&remain[i])
{
has_find=
true;
cur[now]=i;
last[v[i]]=i;
now=v[i];
break;
}
if (!has_find)
{
int minn=t-
1;
for (
int i=point[now];i!=-
1;i=nxt[i])
if (remain[i]) minn=min(minn,deep[v[i]]);
if (!(--num[deep[now]]))
break;
++num[deep[now]=minn+
1];
cur[now]=point[now];
if (now!=s) now=v[last[now]^
1];
}
}
}
int main()
{
tot=-
1;
memset(point,-
1,
sizeof(point));
scanf(
"%d",&n);
s=n+
1,t=s+
1,ss=t+
1,tt=ss+
1;
for (
int i=
1;i<=n;++i)
{
scanf(
"%d",&m);
while (m--)
{
scanf(
"%d",&x);
addedge(i,x,inf);
--d[i],++d[x];
}
addedge(s,i,inf);
addedge(i,t,inf);
}
for (
int i=
1;i<=t;++i)
{
if (d[i]>
0) addedge(ss,i,d[i]);
if (d[i]<
0) addedge(i,tt,-d[i]);
}
isap(ss,tt);
addedge(t,s,inf);
isap(ss,tt);
printf(
"%d\n",remain[tot]);
}
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