网络流+最小生成树的最少割边数--How Many to Be Happy?

题意：https://blog.csdn.net/Ratina/article/details/95200594

思路：

首先我们知道最小生成树就是按长度枚举边，能连就连。

那么，如果这条边在最小生成树里，那我们只需要看比它短的边是不是已经使当前的u---v连通，如果连通最少需要切掉几条（边权为1跑最小割）。

所以我们对边排序，枚举边+重构图跑Dinic就行了。

 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
 #include <cstdio>//sprintf islower isupper
 #include <cstdlib>//malloc  exit strcat itoa system("cls")
 #include <iostream>//pair
 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin);
 #include <bitset>
 //#include <map>
 //#include<unordered_map>
 #include <vector>
 #include <stack>
 #include <set>
 #include <string.h>//strstr substr
 #include <string>
 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
 #include <cmath>
 #include <deque>
 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 #include <vector>//emplace_back
 //#include <math.h>
 #include <cassert>
 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
 //******************
 int abss(int a);
 int lowbit(int n);
 int Del_bit_1(int n);
 int maxx(int a,int b);
 int minn(int a,int b);
 double fabss(double a);
 void swapp(int &a,int &b);
 clock_t __STRAT,__END;
 double __TOTALTIME;
 void _MS(){__STRAT=clock();}
 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 //***********************
 #define rint register int
 #define fo(a,b,c) for(rint a=b;a<=c;++a)
 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define pr printf
 #define sc scanf
 #define ls rt<<1
 #define rs rt<<1|1
 typedef vector<int> VI;
 typedef long long ll;
 const double E=2.718281828;
 const double PI=acos(-1.0);
 //const ll INF=(1LL<<60);
 const int inf=(<<);
 const double ESP=1e-;
 const int mod=(int)1e9+;
 const int N=(int)1e3+;
 const int M=(int)5e3+;

 class DINIC
 {
 public:
 //    const int MAXN=10004,MAXWAY=100005;
     int n,way,max_flow,deep[N];
     int tot,head[N],cur[N];
     struct EDGE{
         int to,next;
         int dis;
     }edge[M];
     void Init(int n_)
     {
         tot=-;//因为加反向边要^1,所以要从0开始;
         n=n_;
         max_flow=;
         for(int i=;i<=n_;++i)
             head[i]=-;
     }
     void add(int from,int to,int V)
     {
         //正向
         ++tot;
         edge[tot].to=to;
         edge[tot].dis=V;
         edge[tot].next=head[from];
         head[from]=tot;
         //反向
         swap(from,to);
         ++tot;
         edge[tot].to=to;
         edge[tot].dis=V;
         edge[tot].next=head[from];
         head[from]=tot;
     }
     queue<int>q;
     bool bfs(int s,int t)
     {
         for(int i=;i<=n;++i)
             deep[i]=inf;
         while(!q.empty())q.pop();
         for(int i=;i<=n;++i)cur[i]=head[i];
         deep[s]=;
         q.push(s);

         while(!q.empty())
         {
             int now=q.front();q.pop();
             for(int i=head[now];i!=-;i=edge[i].next)
             {
                 if(deep[edge[i].to]==inf&&edge[i].dis)
                 {
                     deep[edge[i].to]=deep[now]+;
                     q.push(edge[i].to);
                 }
             }
         }
         return deep[t]<inf;
     }
     int dfs(int now,int t,int limit)
     {
         if(!limit||now==t)return limit;
         int flow=,f;
         for(int i=cur[now];i!=-;i=edge[i].next)
         {
             cur[now]=i;
             if(deep[edge[i].to]==deep[now]+&&(f=dfs(edge[i].to,t,min(limit,edge[i].dis))))
             {
                 flow+=f;
                 limit-=f;
                 edge[i].dis-=f;
                 edge[i^].dis+=f;
                 if(!limit)break;
             }
         }
         return flow;
     }
     void Dinic(int s,int t)
     {
         while(bfs(s,t))
             max_flow+=dfs(s,t,inf);
     }
 }G;
 struct EDGE
 {
     int u,v;
     int val;
     friend bool operator<(EDGE a,EDGE b)
     {
         return a.val<b.val;
     }
 }edge[M];

 int main()
 {
     int n,m;
     sc("%d%d",&n,&m);
     for(int i=;i<=m;++i)
         sc("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].val);
     sort(edge+,edge++m);
     int ans=;
     for(int i=;i<=m;++i)
     {
         G.Init(n);
         for(int j=;j<i;++j)
         {
             if(edge[j].val<edge[i].val)
                 G.add(edge[j].u,edge[j].v,);
         }
         G.Dinic(edge[i].u,edge[i].v);
         ans+=G.max_flow;
     }
     pr("%d\n",ans);
     return ;
 }

 /**************************************************************************************/

 int maxx(int a,int b)
 {
     return a>b?a:b;
 }

 void swapp(int &a,int &b)
 {
     a^=b^=a^=b;
 }

 int lowbit(int n)
 {
     return n&(-n);
 }

 int Del_bit_1(int n)
 {
     return n&(n-);
 }

 int abss(int a)
 {
     return a>?a:-a;
 }

 double fabss(double a)
 {
     return a>?a:-a;
 }

 int minn(int a,int b)
 {
     return a<b?a:b;
 }