最小路径覆盖问题(网络流最大流)

最小路径覆盖问题(网络流最大流)

题目

洛谷题目传送门

题解

网络流题目详讲

code

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iomanip>
#include<algorithm>
#include<ctime>
#include<queue>
#include<stack>
#include<vector>
#define rg register
#define il inline
#define lst long long
#define ldb long double
#define N 550
#define M 100050
using namespace std;
const int Inf=1e9;
il int read()
{
    rg int s=0,m=0;rg char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')m=1;ch=getchar();}
    while(ch>='0'&&ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
    return m?-s:s;
}

int n,m,S,T,ans;
int to[N],tag[N],dep[N];
int hd[N],cur[N],cnt=1;
struct EDGE{int to,nxt,c;}ljl[M<<1];
il void add(rg int p,rg int q,rg int o)
{
    ljl[++cnt]=(EDGE){q,hd[p],o},hd[p]=cnt;
    ljl[++cnt]=(EDGE){p,hd[q],0},hd[q]=cnt;
}

queue<int> Q;
il bool BFS()
{
    for(rg int i=S;i<=T;++i)dep[i]=0;
    while(!Q.empty())Q.pop();
    Q.push(S),dep[S]=1;
    while(!Q.empty())
    {
        rg int now=Q.front();Q.pop();
        for(rg int i=hd[now];i;i=ljl[i].nxt)
        {
            rg int qw=ljl[i].to;
            if(!dep[qw]&&ljl[i].c>0)
            {
                dep[qw]=dep[now]+1;
                Q.push(qw);
            }
        }
    }
    return dep[T];
}

int dfs(rg int now,rg int aim,rg int flow)
{
    if(now==aim)return flow;
    for(rg int &i=cur[now];i;i=ljl[i].nxt)
    {
        rg int qw=ljl[i].to;
        if(ljl[i].c>0&&dep[qw]==dep[now]+1)
        {
            rg int kk=dfs(qw,aim,min(flow,ljl[i].c));
            if(kk>0)
            {
                to[now]=qw;
                if(now!=S)tag[qw-n]=1;
                ljl[i].c-=kk,ljl[i^1].c+=kk;
                return kk;
            }
        }
    }
    return 0;
}

il int Dinic()
{
    rg int ans=0;
    while(BFS())
    {
        for(rg int i=S;i<=T;++i)cur[i]=hd[i];
        while(int kk=dfs(S,T,Inf))ans+=kk;
    }
    for(rg int i=1;i<=n;++i)
        if(!tag[i])
        {
            rg int now=i;
            printf("%d ",now);
            while(to[now]&&to[now]!=T)
            {
                printf("%d ",to[now]-n);
                now=to[now]-n;
            }
            puts("");
        }
    return ans;
}

int main()
{
    n=read(),m=read();
    S=0,T=2*n+1;
    for(rg int i=1;i<=n;++i)
        add(S,i,1),add(i+n,T,1);
    for(rg int i=1;i<=m;++i)
    {
        rg int p=read(),q=read();
        add(p,q+n,1);
    }
    printf("%d\n",n-Dinic());
    return 0;
}