gym-101667E How Many to Be Happy

How Many to Be Happy?

最小割

因为是最小生成树，因此可以考虑对于一条边来说，他的左右两端的点视为处于两个不同的集合，然后只通过该边进行连接，这样最小生成树就必然会利用这条边

比该边大的边显然不用考虑，就考虑比该边边权小的边，然后进行最小割，边流量为 \(1\)（分割成两个集合，且割的边数最少）

#include <iostream>
#include <cstdio>
#include <vector>
#include <array>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 2e4 + 10;
const int inf = 1e8;
int head[maxn], nexx[maxn], to[maxn], vol[maxn], vp = 1;
int dep[maxn], cur[maxn], n, m, s, t;

inline void add(int u, int v, int t)
{
    vp++;
    nexx[vp] = head[u];
    to[vp] = v;
    vol[vp] = t;
    head[u] = vp;
}

bool bfs()
{
    queue<int>q;
    q.push(s);
    for(int i=0; i<=n; i++) cur[i] = head[i];
    for(int i=0; i<=n; i++) dep[i] = 0;
    dep[s] = 1;
    while(q.size())
    {
        int now = q.front();
        q.pop();
        for(int i=head[now]; i; i=nexx[i])
        {
            int nex = to[i];
            if(dep[nex] == 0 && vol[i] > 0)
            {
                dep[nex] = dep[now] + 1;
                q.push(nex);
            }
        }
    }
    return dep[t];
}

int dfs(int now, int flow = inf)
{
    if(now == t) return flow;
    int ans = 0;
    for(int i=head[now]; i && flow; i=nexx[i])
    {
        int nex = to[i];
        if(dep[nex] == dep[now] + 1 && vol[i])
        {
            int f = dfs(nex, min(flow, vol[i]));
            vol[i] -= f;
            vol[i ^ 1] += f;
            ans += f;
            flow -= f;
        }
    }
    return ans;
}

int dinic()
{
    int ans = 0;
    while(bfs())
        ans += dfs(s);
    return ans;
}

int main()
{
    cin >> n >> m;
    vector<array<int, 3> >num(m);
    for(int i=0; i<m; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        num[i] = {c, a, b};
    }
    sort(num.begin(), num.end());
    ll ans = 0;
    for(auto [c, a, b] : num)
    {
        for(int i=0; i<=n; i++) head[i] = 0;
        vp = 1;
        for(auto [cc, aa, bb] : num)
        {
            if(cc >= c) break;
            add(aa, bb, 1);
            add(bb, aa, 0);
            add(bb, aa, 1);
            add(aa, bb, 0);
        }
        s = a;
        t = b;
        ans += dinic();
    }
    cout << ans << endl;
    return 0;
}