P2764 最小路径覆盖问题

P2764 最小路径覆盖问题

有向无环图的最小路径覆盖=点数-二分图最大匹配数

由于之前写二分图最大匹配已经写的很多了，这次换做最大流的做法做一遍

建图：这题的建图也很简单，$[1,n]$的每一个点都构造一个复制，编号为$[n+1,2n]$,左右各一个超级源点，编号为0和$2n+1$，边的容量为1

至于输出路径，一开始想了好久，发现怎么都不对，原来是没有把dinic写在前面，所以相当于一直在假的网络流上跑

一个next数组记录下每一个点的下一个点，并记录下哪些是深度为0的，从这些深度为0的开始遍历

代码：

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e6+7;
const int maxm=2e6+7;
const int INF=1e9;
const ll INFF=1e18;
struct Edge
{
    int to,next,cap,flow;
}edge[maxm];
int tol,head[maxn];
void init()
{
    tol=2;
    mem(head,-1);
}
void add(int u,int v,int w,int rw=0)
{
    edge[tol].to=v;edge[tol].cap=w;edge[tol].flow=0;
    edge[tol].next=head[u];head[u]=tol++;
    edge[tol].to=u;edge[tol].cap=rw;edge[tol].flow=0;
    edge[tol].next=head[v];head[v]=tol++;
}
int Q[maxn],dep[maxn],cur[maxn],sta[maxn],n,m,a,b,D[maxn],next[maxn];
bool bfs(int sx,int ex)
{
    int front=0,tail=0;
    mem(dep,-1);
    dep[sx]=0;
    Q[tail++]=sx;
    while(front<tail)
    {
        int u=Q[front++];
        for (int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if (edge[i].cap>edge[i].flow&&dep[v]==-1)
            {
                dep[v]=dep[u]+1;
                if (v==ex)return true;
                Q[tail++]=v;
            }
        }
    }
    return false;
}
int dinic(int sx,int ex,int n)
{
    int maxflow=0;
    while(bfs(sx,ex))
    {
        rep(i,0,2*n+1)cur[i]=head[i];
        int u=sx,tail=0;
        while(cur[sx]!=-1)
        {
            if (u==ex)
            {
                int tp=INF;
                for (int i=tail-1;i>=0;i--)
                {
                    tp=min(tp,edge[sta[i]].cap-edge[sta[i]].flow);
                }
                maxflow+=tp;
                for (int i=tail-1;i>=0;i--)
                {
                    edge[sta[i]].flow+=tp;
                    edge[sta[i]^1].flow-=tp;
                    if (edge[sta[i]].cap-edge[sta[i]].flow==0)tail=i;
                }
                u=edge[sta[tail]^1].to;
            }
            else if (cur[u]!=-1&&edge[cur[u]].cap>edge[cur[u]].flow&&dep[u]+1==dep[edge[cur[u]].to])
            {
                sta[tail++]=cur[u];
                u=edge[cur[u]].to;
            }
            else
            {
                while(u!=sx&&cur[u]==-1)
                {
                    u=edge[sta[--tail]^1].to;
                }
                cur[u]=edge[cur[u]].next;
            }
        }
    }
    return maxflow;
}
int main()
{
    init();
    scanf("%d%d",&n,&m);
    rep(i,1,m)
    {
        scanf("%d%d",&a,&b);
        add(a,b+n,1,0);
    }
    rep(i,1,n)
    {
        add(0,i,1,0);
        add(i+n,2*n+1,1,0);
    }
    int ans=dinic(0,2*n+1,n);
    rep(u,1,n)
    {
        for (int i=head[u];i!=-1;i=edge[i].next)
        {
            int to=edge[i].to,flow=edge[i^1].flow;
            if (to>n&&flow)
            {
                D[to-n]++;
                next[u]=to-n;
            }
        }
    }
    rep(u,1,n)
    {
        if (!D[u])
        {
            printf("%d ",u);
            for (int i=next[u];i!=0;i=next[i])printf("%d ",i);
            printf("\n");
        }
    }
    W(n-ans);
}

来源：CSDN

作者：w_udixixi

链接：https://blog.csdn.net/w_udixixi/article/details/104459504