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;
}