LibreOJ 6002 最小路径覆盖(最大流)

题解:最小路径覆盖=总点数减去最大匹配数,拆点,按照每条边前一个点连源点,后一个点连汇点跑最大流,即可跑出最大匹配数,然后减一减就可以了~

代码如下:

#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
#define hi puts("hi");
using namespace std;

int head[],next[],v[],w[],deep[];
int s,t,cnt;

void init()
{
    cnt=-;
    memset(head,-,sizeof(head));
    memset(next,-,sizeof(next));
}

void add(int from,int to,int cost)
{
    cnt++;
    next[cnt]=head[from];
    w[cnt]=cost;
    v[cnt]=to;
    head[from]=cnt;
}

void add_edge(int from,int to,int cost)
{
    add(from,to,cost);
    add(to,from,);
}

int bfs(int s,int t)
{
    queue<int> q;
    memset(deep,,sizeof(deep));
    deep[s]=;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-; i=next[i])
        {
            if(w[i]>&&(!deep[v[i]]))
            {
                deep[v[i]]=deep[u]+;
                q.push(v[i]); //
            }
        }
    }
    if(!deep[t])
    {
        return ;
    }
    return ;
}

int dfs(int u,int t,int dist)
{
    if(u==t)
    {
        return dist;
    }
    for(int i=head[u]; i!=-; i=next[i])
    {
        if(w[i]&&(deep[v[i]]==deep[u]+))
        {
            int di=dfs(v[i],t,min(w[i],dist));
            if(di>)
            {
                w[i]-=di;
                w[i^]+=di;
                return di;
            }
        }
    }
    return ;
}

int dinic(int s,int t)
{
    int res=;
    while(bfs(s,t))
    {
        while(int d=dfs(s,t,inf))
        {
            res+=d;
        }
    }
    return res;
}

int n,m;
int vis[];

int solve(int x,int &f)
{
    int loc=x+n;
    vis[x]=;
    for(int i=head[loc];i!=-;i=next[i])
    {
        if(w[i]==&&v[i]!=n*+)
        {
            solve(v[i],f);
        }
    }
    if(f==)
    {
        f=;
    }
    else
    {
        putchar(' ');
    }
    printf("%d",x);
}

int main()
{
    scanf("%d%d",&n,&m);
    init();
    s=;t=n*+;
    for(int i=;i<=n;i++)
    {
        add_edge(s,i,);
        add_edge(i+n,t,);
    }
    int x,y;
    while(m--)
    {
        scanf("%d%d",&x,&y);
        add_edge(x,y+n,);
    }
    int ans=n-dinic(s,t);
    for(int i=head[t];i!=-;i=next[i])
    {
        if(w[i]==&&!vis[v[i]-n])
        {
            int f=;
            solve(v[i]-n,f);
            puts("");
        }
    }
    printf("%d\n",ans);
}