洛谷

https://www.luogu.org/problemnew/solution/P2762

最小割对应的点，在最后一次更新中dinic的bfs会把他的dep重置掉。所以可以根据这个性质复原最小割。

#include<bits/stdc++.h>
using namespace std;
#define ll long long

/* dinic begin */

const int MAXN=10100;
const int MAXM=100010;
const int INF=0x3f3f3f3f;
struct Edge{
    int to,next,cap,flow;
}edge[MAXM];

int tol;
int head[MAXN];

void init(){
    tol=2;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int w){
    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=0;edge[tol].flow=0;
    edge[tol].next=head[v];head[v]=tol++;
}

int Q[MAXN];
int dep[MAXN],cur[MAXN],sta[MAXN];
bool bfs(int s,int t,int n){
    int front=0,tail=0;
    memset(dep,-1,sizeof(dep[0])*(n+1));
    dep[s]=0;
    Q[tail++]=s;
    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==t)
                    return true;
                Q[tail++]=v;
            }
        }
    }
    return false;
}

int dinic(int s,int t,int n){
    //n最后一个节点的编号的下一个编号
    int maxflow=0;
    while(bfs(s,t,n)){
        for(int i=0;i<n;i++)cur[i]=head[i];
        int u=s,tail=0;
        while(cur[s]!=-1){
            if(u==t){
                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!=s&&cur[u]==-1){
                    u=edge[sta[--tail]^1].to;
                }
                cur[u]=edge[cur[u]].next;
            }
        }
    }
    return maxflow;
}

/* dinic end */


int m,n;
int main(){
    init();
    scanf("%d%d",&m,&n);
    char buf[40000];
    fgets(buf,40000,stdin);

    int s=0,t=m+n+1;
    int sum=0;
    for(int i=1;i<=m;i++){
        fgets(buf,40000,stdin);
        stringstream ss(buf);
        int w;
        ss>>w;
        //cout<<w<<endl;
        sum+=w;
        addedge(s,i,w);
        int j;
        while(ss>>j){
            //cout<<j<<endl;
            addedge(i,j+m,INF);
        }
    }

    for(int i=1;i<=n;i++){
        int w;
        scanf("%d",&w);
        //cout<<"w="<<w<<endl;
        addedge(i+m,t,w);
    }

    int maxflow=dinic(s,t,t);

    vector<int> v1;
    for(int u=1;u<=m;u++){
        if(dep[u]!=-1){
            v1.push_back(u);
            //最后一次dinic还有dep的边说明没有被割掉
        }
    }

    vector<int>v2;
    for(int u=1;u<=n;u++){
        if(dep[u+m]!=-1){
            v2.push_back(u);
            //最后一次dinic还有dep的边说明没有被割掉
        }
    }

    for(int i=0;i<v1.size();i++){
        printf("%d%c",v1[i]," 
"[i==v1.size()-1]);
    }

    for(int i=0;i<v2.size();i++){
        printf("%d%c",v2[i]," 
"[i==v2.size()-1]);
    }


    printf("%d
",sum-maxflow);

}