追查坏牛奶(最大流)

Posted 2021-11-23 hale522520

进入今天的正题，追查坏牛奶

思想的话不会人家的玄学求最小割的边数，于是自己想了个神奇的乱搞，先求出最大流，即最小割，然后枚举每条边看是否完全属于割集，然后将其永久去掉

然后将整个输出即可

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=100010;
const ll INF=0x7ffffffff;
ll m,n,s,t,pre[N],head[N],cnt=1,last[N],dep[N],tot,sum[N],ans,Ans,ff;
struct edge
{ ll from,to,nx,flow;} e[N];
struct node
{ ll x,y,z;} kk[N];
void add_edge(ll from,ll to,ll flow)
{ cnt++;e[cnt].from=from;e[cnt].flow=flow;
  e[cnt].nx=head[from];e[cnt].to=to;head[from]=cnt;
  cnt++;e[cnt].flow=0;e[cnt].to=from;
  e[cnt].from=to;e[cnt].nx=head[to];head[to]=cnt;
}
bool bfs(int s,int t)
{ memset(dep,0,sizeof(dep));
  queue<ll> que;dep[s]=1;que.push(s);
  while (!que.empty())
  { ll x=que.front();que.pop();
    for (ll i=head[x];i;i=e[i].nx)
    { ll y=e[i].to;
      if (dep[y]==0&&e[i].flow>0)
      { dep[y]=dep[x]+1;
        que.push(y);
      }
    }
  }
 if (dep[t]==0) return false;
 else return true;  
}
ll dfs(int x,ll limit,int t)
{ if(x==t) return limit;
  ll used=0;
  for (ll i=head[x];i;i=e[i].nx)
  { ll y=e[i].to;
    if (dep[y]==dep[x]+1&&e[i].flow>0)
    { ll di=dfs(y,min(limit-used,e[i].flow),t);
      if (di>0)
      { e[i].flow-=di;
        e[i^1].flow+=di;
        used+=di;
        if (used==limit) return used;
      }
    }
  }
 if (!used) dep[x]=-2;
 return used; 
}
void dinic()
{ while (bfs(1,n)) ans+=dfs(1,INF,n);}
void cs()
{  ans=0;cnt=1;memset(head,0,sizeof(head));
   for (int j=1;j<=m;j++) add_edge(kk[j].x,kk[j].y,kk[j].z);
   for (int k=1;k<=tot;k++) {e[sum[k]*2].flow=0;} 
}
int main()
{ 
  scanf("%lld%lld",&n,&m);
  for (ll i=1;i<=m;i++)
  { ll x,y,z;
    scanf("%lld%lld%lld",&x,&y,&z);
    kk[i].x=x;kk[i].y=y;kk[i].z=z;
    add_edge(x,y,z);
  }
  dinic();ff=ans;Ans=ans;
  for (int i=1;i<=m;i++)
  { cs();
    ll f=e[i*2].flow;
    e[i*2].flow=0;
    dinic();
    if (ans==0&&f==ff) {tot=1;sum[tot]=i;break;}
    if (ans+f==Ans) {tot++;sum[tot]=i;Ans=ans;}
  }
  printf("%lld %lld ",ff,tot);
  for (int i=1;i<=tot;i++)
  printf("%lld ",sum[i]);
  return 0;
}