P1344 [USACO4.4] 追查坏牛奶 Pollutant Control

Description

给定一张 $n$ 点 $m$ 边的有向图，求 $1\to n$ 的最小割的容量和边数。

Analysis

第一问是模板，直接套最小割（最大流）即可。
第二问的话，我们可以将流满的边容量改成 $1$，没流满的改成 $\infty$，在新图上跑一遍最小割即可求出。

Code

最大流贴的是 jiangly 的模板

// Problem: P1344 [USACO4.4] 追查坏牛奶 Pollutant Control
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P1344
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;

using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;

template<class T>
bool chmax(T &a, const T &b){
	if(a < b){ a = b; return true; }
	return false;
}

template<class T>
bool chmin(T &a, const T &b){
	if(a > b){ a = b; return true; }
	return false;
}

const int INF = 2e9;

template<class T>
struct MaxFlow {
    struct _Edge {
        int to;
        T cap;
        _Edge(int to, T cap) : to(to), cap(cap) {}
    };
    
    int n;
    T inf;
    std::vector<_Edge> e;
    std::vector<std::vector<int>> g;
    std::vector<int> cur, h;
    
    MaxFlow() {}
    MaxFlow(int n) {
        init(n);
        inf = numeric_limits<T>::max();
    }
    
    void init(int n) {
        this->n = n;
        e.clear();
        g.assign(n, {});
        cur.resize(n);
        h.resize(n);
    }
    
    bool bfs(int s, int t) {
        h.assign(n, -1);
        std::queue<int> que;
        h[s] = 0;
        que.push(s);
        while (!que.empty()) {
            const int u = que.front();
            que.pop();
            for (int i : g[u]) {
                int v = e[i].to;
                T c = e[i].cap;
                if (c > 0 && h[v] == -1) {
                    h[v] = h[u] + 1;
                    if (v == t) {
                        return true;
                    }
                    que.push(v);
                }
            }
        }
        return false;
    }
    
    T dfs(int u, int t, T f) {
        if (u == t) {
            return f;
        }
        auto r = f;
        for (int &i = cur[u]; i < int(g[u].size()); ++i) {
            const int j = g[u][i];
            int v = e[j].to;
            T c = e[j].cap;
            if (c > 0 && h[v] == h[u] + 1) {
                auto a = dfs(v, t, std::min(r, c));
                e[j].cap -= a;
                e[j ^ 1].cap += a;
                r -= a;
                if (r == 0) {
                    return f;
                }
            }
        }
        return f - r;
    }
    void addEdge(int u, int v, T c) {
        g[u].push_back(e.size());
        e.emplace_back(v, c);
        g[v].push_back(e.size());
        e.emplace_back(u, 0);
    }
    T flow(int s, int t) {
        T ans = 0;
        while (bfs(s, t)) {
            cur.assign(n, 0);
            ans += dfs(s, t, std::numeric_limits<T>::max());
        }
        return ans;
    }
    
    std::vector<bool> minCut() {
        std::vector<bool> c(n);
        for (int i = 0; i < n; i++) {
            c[i] = (h[i] != -1);
        }
        return c;
    }
    
    struct Edge {
        int from;
        int to;
        T cap;
        T flow;
    };
    std::vector<Edge> edges() {
        std::vector<Edge> a;
        for (int i = 0; i < e.size(); i += 2) {
            Edge x;
            x.from = e[i + 1].to;
            x.to = e[i].to;
            x.cap = e[i].cap + e[i + 1].cap;
            x.flow = e[i + 1].cap;
            a.push_back(x);
        }
        return a;
    }
};


signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n, m;
	cin >> n >> m;
	
	// Find the capacity of the mincut.
	MaxFlow<int> G(n);
	for(int i = 0, u, v, w; i < m; i++){
	    cin >> u >> v >> w;
	    u--, v--;
	    G.addEdge(u, v, w);
	}
	
	int ans = G.flow(0, n - 1);
	
    // Find the number of edges of the mincut.
	auto edges = G.edges();

    // Build a new network graph.
	MaxFlow<int> G2(n);
	for(auto &edge: edges){
	    if(edge.cap == edge.flow) G2.addEdge(edge.from, edge.to, 1);
	    else G2.addEdge(edge.from, edge.to, INF);
	}
	int ecnt = G2.flow(0, n - 1);
	
	cout << ans << ' ' << ecnt << endl;
	return 0;
}