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=(1<<30);
const double ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e3+10;
const int M=(int)5e3+5;

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=-1;//因为加反向边要^1,所以要从0开始;
        n=n_;
        max_flow=0;
        for(int i=0;i<=n_;++i)
            head[i]=-1;
    }
    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=1;i<=n;++i)
            deep[i]=inf;
        while(!q.empty())q.pop();
        for(int i=1;i<=n;++i)cur[i]=head[i];
        deep[s]=0;
        q.push(s);

        while(!q.empty())
        {
            int now=q.front();q.pop();
            for(int i=head[now];i!=-1;i=edge[i].next)
            {
                if(deep[edge[i].to]==inf&&edge[i].dis)
                {
                    deep[edge[i].to]=deep[now]+1;
                    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=0,f;
        for(int i=cur[now];i!=-1;i=edge[i].next)
        {
            cur[now]=i;
            if(deep[edge[i].to]==deep[now]+1&&(f=dfs(edge[i].to,t,min(limit,edge[i].dis))))
            {
                flow+=f;
                limit-=f;
                edge[i].dis-=f;
                edge[i^1].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=1;i<=m;++i)
        sc("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].val);
    sort(edge+1,edge+1+m);
    int ans=0;
    for(int i=1;i<=m;++i)
    {
        G.Init(n);
        for(int j=1;j<i;++j)
        {
            if(edge[j].val<edge[i].val)
                G.add(edge[j].u,edge[j].v,1);
        }
        G.Dinic(edge[i].u,edge[i].v);
        ans+=G.max_flow;
    }
    pr("%d
",ans);
    return 0;
}

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

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-1);
}

int abss(int a)
{
    return a>0?a:-a;
}

double fabss(double a)
{
    return a>0?a:-a;
}

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