LibreOJ 6002 最小路径覆盖(最大流)

Posted 2020-10-27 Styx-ferryman

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[100010],next[100010],v[100010],w[100010],deep[100010];
int s,t,cnt;

void init()
{
    cnt=-1;
    memset(head,-1,sizeof(head));
    memset(next,-1,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,0);
}

int bfs(int s,int t)
{
    queue<int> q;
    memset(deep,0,sizeof(deep));
    deep[s]=1;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1; i=next[i])
        {
            if(w[i]>0&&(!deep[v[i]]))
            {
                deep[v[i]]=deep[u]+1;
                q.push(v[i]); //
            }
        }
    }
    if(!deep[t])
    {
        return 0;
    }
    return 1;
}

int dfs(int u,int t,int dist)
{
    if(u==t)
    {
        return dist;
    }
    for(int i=head[u]; i!=-1; i=next[i])
    {
        if(w[i]&&(deep[v[i]]==deep[u]+1))
        {
            int di=dfs(v[i],t,min(w[i],dist));
            if(di>0)
            {
                w[i]-=di;
                w[i^1]+=di;
                return di;
            }
        }
    }
    return 0;
}

int dinic(int s,int t)
{
    int res=0;
    while(bfs(s,t))
    {
        while(int d=dfs(s,t,inf))
        {
            res+=d;
        }
    }
    return res;
}

int n,m;
int vis[100010];

int solve(int x,int &f)
{
    int loc=x+n;
    vis[x]=1;
    for(int i=head[loc];i!=-1;i=next[i])
    {
        if(w[i]==1&&v[i]!=n*2+1)
        {
            solve(v[i],f);
        }
    }
    if(f==1)
    {
        f=0;
    }
    else
    {
        putchar(\' \');
    }
    printf("%d",x);
}

int main()
{
    scanf("%d%d",&n,&m);
    init();
    s=0;t=n*2+1;
    for(int i=1;i<=n;i++)
    {
        add_edge(s,i,1);
        add_edge(i+n,t,1);
    }
    int x,y;
    while(m--)
    {
        scanf("%d%d",&x,&y);
        add_edge(x,y+n,1);
    }
    int ans=n-dinic(s,t);
    for(int i=head[t];i!=-1;i=next[i])
    {
        if(w[i]==1&&!vis[v[i]-n])
        {
            int f=1;
            solve(v[i]-n,f);
            puts("");
        }
    }
    printf("%d\\n",ans);
}