【loj】#6001. 「网络流 24 题」太空飞行计划(最大权闭合子图)

记录一个菜逼的成长。。

题目链接
hihocoder

used数组中保存的就是最小割点集
$ans = \sum 正权值 - 最小割容量$

#include <bits/stdc++.h>
using namespace std;
#define ALL(v) (v).begin(),(v).end()
#define cl(a,b) memset(a,b,sizeof(a))
#define clr clear()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define fin freopen("D://in.txt","r",stdin)
#define fout freopen("D://out.txt","w",stdout)
#define lson t<<1,l,mid
#define rson t<<1|1,mid+1,r
#define ls t<<1
#define rs ls|1
#define seglen(t) (node[t].r-node[t].l+1)
#define lowbit(x) (x)&(-x)
typedef long long LL;
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int MAX_V = 100 + 10;
struct edge{
  int to,cap,rev;
  edge(){}
  edge(int _to,int _cap,int _rev):to(_to),cap(_cap),rev(_rev){}
};
vector<edge>G[MAX_V];
bool used[MAX_V];
void add_edge(int from,int to,int cap)
{
  G[from].push_back(edge(to,cap,G[to].size()));
  G[to].push_back(edge(from,0,G[from].size()-1));
}
int dfs(int v,int t,int f)
{
  if( v == t )return f;
  used[v] = true;
  for( int i = 0; i < G[v].size(); i++ ){
    edge &e = G[v][i];
    if(!used[e.to] && e.cap > 0){
      int d = dfs(e.to,t,min(f,e.cap));
      if(d > 0){
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
int max_flow(int s,int t)
{
  int flow = 0;
  for(;;){
    memset(used,0,sizeof(used));
    int f = dfs(s,t,INF);
    if(!f)return flow;
    flow += f;
  }
}
int main()
{
  int n,m;
  scanf("%d%d\n",&n,&m);
  int s = 0,t = n + m + 1;
  int sum = 0,k;
  for( int i = 1; i <= n; i++ ){
    string str;
    getline(cin,str);
    stringstream ss(str);
    ss >> k;
    sum += k;
    add_edge(s,i,k);
    while(ss >> k){
      add_edge(i,k+n,INF);
    }
  }
  for( int i = 1; i <= m; i++ ){
    int x;scanf("%d",&x);
    add_edge(i+n,t,x);
  }
  int ret = max_flow(s,t);
  for( int i = 1; i <= n; i++ ){
    if(used[i])printf("%d ",i);
  }
  puts("");
  for( int i = n+1; i <= m+n; i++ ){
    if(used[i])printf("%d ",i-n);
  }
  puts("");
  printf("%d\n",sum - ret);
  return 0;
}